# Class 11 RD Sharma Solutions – Chapter 11 Trigonometric Equations – Exercise 11.1

Last Updated : 28 Apr, 2021

### (i) sin Î¸ = 1/2

Solution:

We are given,

=> sin Î¸ = 1/2

=> sin Î¸ = sin Ï€/6

We know if sin Î¸ = sin a, the general solution is given by, Î¸ = nÏ€ + (âˆ’1)n a; n âˆˆ Z.

=> Î¸ = nÏ€ + (âˆ’1)n (Ï€/6) ; n âˆˆ Z

### (ii) cos Î¸ = âˆ’âˆš3/2

Solution:

We are given,

=> cos Î¸ = âˆ’âˆš3/2

=> cos Î¸ = cos (Ï€ + Ï€/6)

=> cos Î¸ = cos (7Ï€/6)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a ; n âˆˆ  Z.

=> Î¸ = 2nÏ€ Â± (7Ï€/6) ; n âˆˆ  Z

### (iii) cosec Î¸ = âˆ’âˆš2

Solution:

We are given,

=> cosec Î¸ = âˆ’âˆš2

=> 1/sin Î¸ = âˆ’âˆš2

=> sin Î¸ = âˆ’1/âˆš2

=> sin Î¸ = âˆ’sin (Ï€/4)

As sin (âˆ’Î¸) = âˆ’ sin Î¸, we have,

=> sin Î¸ = sin (âˆ’Ï€/4)

We know if sin Î¸ = sin a, the general solution is given by, Î¸ = nÏ€ + (âˆ’1)n a ; n âˆˆ  Z.

=> Î¸ = nÏ€ + (âˆ’1)n+1 (Ï€/4) ; n âˆˆ  Z

### (iv) sec Î¸ = âˆš2

Solution:

We are given,

=> sec Î¸ = âˆš2

=> 1/sec Î¸ = âˆš2

=> cos Î¸ = 1/âˆš2

=> cos Î¸ = cos Ï€/4

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a ; n âˆˆ  Z.

=> Î¸ = 2nÏ€ Â± (Ï€/4) ; n âˆˆ  Z

### (v) tan Î¸ = âˆ’1/âˆš3

Solution:

We are given,

=> tan Î¸ = âˆ’1/âˆš3

=> tan Î¸ = âˆ’tan (Ï€/6)

As tan (âˆ’Î¸) = âˆ’ tan Î¸, we have,

=> tan Î¸ = tan (âˆ’Ï€/6)

We know if tan Î¸ = tan a, the general solution is given by, Î¸ = nÏ€ + a ; n âˆˆ  Z.

=> Î¸ = nÏ€ âˆ’ (Ï€/6) ; n âˆˆ  Z

### (vi) âˆš3 sec Î¸ = 2

Solution:

We are given,

=> âˆš3 sec Î¸ = 2

=> âˆš3 (1/cos Î¸) = 2

=> cos Î¸ = âˆš3/2

=> cos Î¸ = cos (Ï€/6)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a; n âˆˆ Z.

=> Î¸ = 2nÏ€ Â± (Ï€/6) ; n âˆˆ Z

### (i) sin 2Î¸ = âˆš3/2

Solution:

We are given,

=> sin 2Î¸ = âˆš3/2

=> sin 2Î¸ = sin (Ï€/3)

We know if sin Î¸ = sin a, the general solution is given by, Î¸ = nÏ€ + (âˆ’1)n a; n âˆˆ Z.

=> 2Î¸ = nÏ€ + (âˆ’1)n (Ï€/3)

=> Î¸ = nÏ€/2 + (âˆ’1)n (Ï€/6) ; n âˆˆ Z

### (ii) cos 3Î¸ = 1/2

Solution:

We are given,

=> cos 3Î¸ = 1/2

=> cos 3Î¸ = cos (Ï€/3)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a; n âˆˆ Z.

=> 3Î¸ = 2nÏ€ Â± (Ï€/3)

=> Î¸ = 2nÏ€/3 Â± (Ï€/9) ; n âˆˆ Z

### (iii) sin 9Î¸ = sin Î¸

Solution:

We are given,

=> sin 9Î¸ = sin Î¸

=> sin 9Î¸ â€“ sin Î¸ = 0

=> 2 cos 5Î¸ sin 4Î¸ = 0

=> cos 5Î¸ = 0 or sin 4Î¸ = 0

We know, if cos Î¸ = 0, then Î¸ = (2n+1)Ï€/2 and if sin Î¸ = 0, then Î¸ = nÏ€ where n âˆˆ  Z.

=> 5Î¸ = (2n+1)Ï€/2 or 4Î¸ = nÏ€

=> Î¸ = (2n+1)Ï€/10 or Î¸ = nÏ€/4, n âˆˆ  Z

### (iv) sin 2Î¸ = cos 3Î¸

Solution:

We are given,

=> sin 2Î¸ = cos 3Î¸

=> cos 3Î¸ = sin 2Î¸

=> cos 3Î¸ = cos (Ï€/2 âˆ’ 2Î¸)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a ; n âˆˆ  Z.

=> 3Î¸ = 2nÏ€ Â± (Ï€/2 âˆ’ 2Î¸)

=> 3Î¸ = 2nÏ€ + Ï€/2 âˆ’ 2Î¸ or 3Î¸ = 2nÏ€ âˆ’ Ï€/2 + 2Î¸

=> 5Î¸ = 2nÏ€ + Ï€/2 or Î¸ = 2nÏ€ âˆ’ Ï€/2

=> 5Î¸ = (4n+1) (Ï€/2) or Î¸ = (4nâˆ’1) (Ï€/2)

=> Î¸ = (4n+1)Ï€/10 or Î¸ = (4nâˆ’1)Ï€/2, n âˆˆ  Z

### (v) tan Î¸ + cot 2Î¸ = 0

Solution:

We are given,

=> tan Î¸ + cot 2Î¸ = 0

=> cot 2Î¸ = âˆ’ tan Î¸

=> tan 2Î¸ = âˆ’ cot Î¸

=> tan 2Î¸ = âˆ’ tan (Ï€/2 âˆ’ Î¸)

As tan (âˆ’Î¸) = âˆ’ tan Î¸, we have,

=> tan 2Î¸ = tan (Î¸ âˆ’ Ï€/2)

We know if tan Î¸ = tan a, the general solution is given by, Î¸ = nÏ€ + a ; n âˆˆ  Z.

=> 2Î¸ = nÏ€ + Î¸ âˆ’ Ï€/2

=> Î¸ = nÏ€ âˆ’ Ï€/2 ; n âˆˆ  Z

### (vi) tan 3Î¸ = cot Î¸

Solution:

We are given,

=> tan 3Î¸ = cot Î¸

=> tan 3Î¸ = tan (Ï€/2 âˆ’ Î¸)

We know if tan Î¸ = tan a, the general solution is given by, Î¸ = nÏ€ + a ; n âˆˆ  Z.

=> 3Î¸ = nÏ€ + Ï€/2 âˆ’ Î¸

=> 4Î¸ = nÏ€ + Ï€/2

=> Î¸ = nÏ€/4 + Ï€/8 ; n âˆˆ  Z

### (vii) tan 2Î¸ tan Î¸ = 1

Solution:

We are given,

=> tan 2Î¸ tan Î¸ = 1

=> tan 2Î¸ = cot Î¸

=> tan 2Î¸ = tan (Ï€/2 âˆ’ Î¸)

We know if tan Î¸ = tan a, the general solution is given by, Î¸ = nÏ€ + a ; n âˆˆ  Z.

=> 2Î¸ = nÏ€ + Ï€/2 âˆ’ Î¸

=> 3Î¸ = nÏ€ + Ï€/2

=> Î¸ = nÏ€/3 + Ï€/6 ; n âˆˆ  Z

### (viii) tan mÎ¸ + cot nÎ¸ = 0

Solution:

We are given,

=> tan mÎ¸ + cot nÎ¸ = 0

=> sin mÎ¸/cos mÎ¸ + cos nÎ¸/sin nÎ¸ = 0

=> sin mÎ¸ sin nÎ¸ + cos nÎ¸ cos mÎ¸ = 0

=> cos (mâˆ’n)Î¸ = 0

We know, if cos Î¸ = 0, then Î¸ = (2k+1)Ï€/2 where k âˆˆ  Z.

=> (mâˆ’n)Î¸ = (2k+1)Ï€/2

=> Î¸ = (2k+1)Ï€/2(mâˆ’n) ; n âˆˆ  Z

### (ix) tan pÎ¸ = cot qÎ¸

Solution:

We are given,

=> tan pÎ¸ = cot qÎ¸

=> tan pÎ¸ = tan (Ï€/2 âˆ’ qÎ¸)

We know if tan Î¸ = tan a, the general solution is given by, Î¸ = nÏ€ + a ; n âˆˆ  Z.

=> pÎ¸ = nÏ€ + Ï€/2 âˆ’ qÎ¸

=> (p + q)Î¸ = (2n+1)Ï€/2

=> Î¸ = (2n+1)Ï€/2(p + q) ; n âˆˆ  Z

### (x) sin 2x + cos x = 0

Solution:

We are given,

=> sin 2x + cos x = 0

=> cosx = âˆ’ sin 2x

=> cos x = cos (Ï€/2 + 2x)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a ; n âˆˆ  Z.

=> x = 2nÏ€ Â± (Ï€/2 + 2x)

=> x = 2nÏ€ + Ï€/2 + 2x or x =  2nÏ€ âˆ’ Ï€/2 âˆ’ 2x

=> x = âˆ’(4n+1)Ï€/2 or 3x = (4nâˆ’1)Ï€/2

=> x = âˆ’(4n+1)Ï€/2 or 3x = (4nâˆ’1)Ï€/2

=> x = (4mâˆ’1)Ï€/2, where m = âˆ’n or x = (4nâˆ’1)Ï€/6, n âˆˆ  Z

### (xi) sin Î¸ = tan Î¸

Solution:

We are given,

=> sin Î¸ = tan Î¸

=> sin Î¸ = sin Î¸/cos Î¸

=> sin Î¸ (cos Î¸ âˆ’ 1) = 0

=> sin Î¸ = 0 or cos Î¸ = 1

=> sin Î¸ = 0 or cos Î¸ = cos 0

We know, if cos Î¸ = 0, then Î¸ = (2n+1)Ï€/2 and if sin Î¸ = 0, then Î¸ = nÏ€ where n âˆˆ  Z.

=> Î¸ = nÏ€ or Î¸ = 2nÏ€, n âˆˆ  Z

### (xii) sin 3x + cos 2x = 0

Solution:

We are given,

=> sin 3x + cos 2x = 0

=> cos 2x = âˆ’sin 3x

=>cos 2x = cos (Ï€/2 + 3x)

We know if cos Î¸ = cos a, the general solution is given by, Î¸ = 2nÏ€ Â± a ; n âˆˆ  Z.

=> 2x = 2nÏ€ Â± (Ï€/2 + 3x)

=> 2x = 2nÏ€ + Ï€/2 + 3x or 2x = 2nÏ€ âˆ’ Ï€/2 âˆ’ 3x

=> âˆ’x = 2nÏ€ + Ï€/2 or 5x = 2nÏ€ âˆ’ Ï€/2

=> x = 2mÏ€ âˆ’ Ï€/2, where m = âˆ’n or x = (2nÏ€ âˆ’ Ï€/2)/5

=> x = (4mâˆ’1)Ï€/2, where m = âˆ’n or x = (4nâˆ’1)Ï€/10, n âˆˆ  Z

### (i) sin2 Î¸ âˆ’ cos Î¸ = 1/4

Solution:

We have,

=> sin2 Î¸ âˆ’ cos Î¸ = 1/4

=> 1 âˆ’ cos2 Î¸ âˆ’ cos Î¸ = 1/4

=> cos2 Î¸ + cos Î¸ âˆ’ 3/4 = 0

=> 4cos2 Î¸ + 4cos Î¸ âˆ’ 3 = 0

=> 4cos2 Î¸ + 6cos Î¸ âˆ’ 2cos Î¸ âˆ’ 3 = 0

=> 2cos Î¸ (2cos + 3) âˆ’ (2cos Î¸ + 3) = 0

=> (2cos Î¸ âˆ’ 1) (2cos Î¸ + 3) = 0

=> cos Î¸ = 1/2 or cos Î¸ = âˆ’3/2

Ignoring cos Î¸ = âˆ’3/2 as âˆ’1 â‰¤ cosÎ¸ â‰¤ 1. So, we have, cos Î¸ = 1/2.

=> cos Î¸ = cos Ï€/3

=> Î¸ = 2nÏ€ Â± Ï€/3 ; n âˆˆ  Z

### (ii) 2cos2 Î¸ âˆ’ 5cos Î¸ + 2 = 0

Solution:

We are given,

=> 2cos2 Î¸ âˆ’ 5cos Î¸ + 2 = 0

=> 2cos2 Î¸ âˆ’ 4cos Î¸ âˆ’ cos Î¸ + 2 = 0

=> 2cos Î¸ (cos Î¸ âˆ’ 2) âˆ’ (cos Î¸ âˆ’ 2) = 0

=> (2cos Î¸ âˆ’ 1) (cos Î¸ âˆ’ 2) = 0

=> cos Î¸ = 1/2 or cos Î¸ = 2

Ignoring cos Î¸ = 2 as âˆ’1 â‰¤ cosÎ¸ â‰¤ 1. So, we have, cos Î¸ = 1/2.

=> cos Î¸ = cos Ï€/3

=> Î¸ = 2nÏ€ Â± Ï€/3 ; n âˆˆ  Z

### (iii) 2sin2 x + âˆš3cos x + 1 = 0

Solution:

We are given,

=> 2sin2 x + âˆš3cos x + 1 = 0

=> 2 (1 âˆ’ cos2 x) + âˆš3cos x + 1 = 0

=> 2cos2 x âˆ’ âˆš3cos x âˆ’ 3 = 0

=> 2cos2 x âˆ’ 2âˆš3cos x + âˆš3cos x âˆ’ 3 = 0

=> 2cosx (cos x âˆ’ âˆš3) + âˆš3(cos x âˆ’ âˆš3) = 0

=> (2cosx + âˆš3) (cos x âˆ’ âˆš3) = 0

=> x = âˆ’âˆš3/2 or x = âˆš3

Ignoring cos x = âˆš3 as âˆ’1 â‰¤ cosÎ¸ â‰¤ 1. So, we have, cos x = âˆ’âˆš3/2.

=> cos x = âˆ’ cos Ï€/6

=> cos x = cos (Ï€ âˆ’ Ï€/6)

=> cos x = cos (5Ï€/6)

=> x = 2nÏ€ Â± 5Ï€/6 ; n âˆˆ  Z

### (iv) 4sin2 Î¸ âˆ’ 8cos Î¸ + 1 = 0

Solution:

We are given,

=> 4sin2 Î¸ âˆ’ 8cos Î¸ + 1 = 0

=> 4 (1 âˆ’ cos2 Î¸) âˆ’ 8cos Î¸ + 1 = 0

=> 4 cos2 Î¸ + 8 cos Î¸ âˆ’ 5 = 0

=> 4 cos2 Î¸ + 10 cos Î¸ âˆ’ 2 cos Î¸ âˆ’ 5 = 0

=> 2cos Î¸ (2cos Î¸ +5) âˆ’ (2cos Î¸ +5) = 0

=> (2cos Î¸ âˆ’ 1) (2cos Î¸ +5) = 0

=> cos Î¸ = 1/2 or cos Î¸ = âˆ’5/2

Ignoring cos x = âˆ’5/2 as âˆ’1 â‰¤ cosÎ¸ â‰¤ 1. So, we have, cos x = 1/2.

=> cos x = cos Ï€/3

=> x = 2nÏ€ Â± Ï€/3 ; n âˆˆ  Z

### (v) tan2 x + (1 âˆ’ âˆš3)tan x âˆ’ âˆš3 = 0

Solution:

We are given,

=> tan2 x + (1 âˆ’ âˆš3)tan x âˆ’ âˆš3 = 0

=> tan2 x + tan x âˆ’ âˆš3tan x âˆ’ âˆš3 = 0

=> tan x (tan x + 1) âˆ’ âˆš3 (tan x + 1) = 0

=> (tan x âˆ’ âˆš3) (tan x + 1) = 0

=> tan x = âˆš3 or tan x = âˆ’1

=> tan x = tan Ï€/3 or tan x = âˆ’tan Ï€/4

=> tan x = tan Ï€/3 or tan x = tan (âˆ’Ï€/4)

=> x = nÏ€ + Ï€/3 or x = nÏ€ âˆ’ Ï€/4, n âˆˆ  Z

### (vi) 3 cos2 Î¸ âˆ’ 2âˆš3 sin Î¸ cos Î¸ âˆ’ 3 sin2 Î¸ = 0

Solution:

We have,

=> 3 cos2 Î¸ âˆ’ 2âˆš3 sin Î¸ cos Î¸ âˆ’ 3 sin2 Î¸ = 0

=> âˆš3 cos2 Î¸ âˆ’ 2 sin Î¸ cos Î¸ âˆ’ âˆš3 sin2 Î¸ = 0

=> âˆš3 cos2 Î¸ + sin Î¸ cos Î¸ âˆ’ 3 sin Î¸ cos Î¸ âˆ’ âˆš3 sin2 Î¸ = 0

=> cos Î¸ (âˆš3cos Î¸ + sin Î¸) âˆ’ âˆš3sin Î¸ (âˆš3cos Î¸ + sin Î¸) = 0

=> (âˆš3cos Î¸ + sin Î¸) (cos Î¸ âˆ’ âˆš3sin Î¸) = 0

=> tan Î¸ = âˆ’âˆš3 or tan Î¸ = 1/âˆš3

=> tan Î¸ = âˆ’ tan Ï€/3 or tan Î¸ = tan Ï€/6

=> tan Î¸ = tan (âˆ’Ï€/3) or tan Î¸ = tan Ï€/6

=> Î¸ = nÏ€ âˆ’ Ï€/3 or Î¸ = nÏ€ + Ï€/6, n âˆˆ  Z

### (vii) cos 4Î¸ = cos 2Î¸

Solution:

We have,

=> cos 4Î¸ = cos 2Î¸

=>  cos 4Î¸ âˆ’ cos 2Î¸ = 0

=> 2 sin 3Î¸ sin Î¸ = 0

=> sin 3Î¸ = 0 or sin Î¸ = 0

=> 3Î¸ = nÏ€ or Î¸ = nÏ€

=> Î¸ = nÏ€/3 or Î¸ = nÏ€, n âˆˆ  Z

### (i) cos Î¸ + cos 2Î¸ + cos 3Î¸ = 0

Solution:

We are given,

=> cos 2Î¸ + (cos Î¸ + cos 3Î¸) = 0

=> cos 2Î¸ + 2cos 2Î¸ cos Î¸ = 0

=> cos 2Î¸ (1 + 2cos Î¸) = 0

=> cos 2Î¸ = 0 or cos Î¸ = âˆ’1/2

=> 2Î¸ = (2n+1)Ï€/2 or cos Î¸ = cos (Ï€ âˆ’ Ï€/3)

=> Î¸ = (2n+1)Ï€/4 or Î¸ = 2nÏ€ Â± (2Ï€/3), n âˆˆ  Z

### (ii) cos Î¸ + cos 3Î¸ âˆ’ cos 2Î¸ = 0

Solution:

We are given,

=> cos Î¸ + cos 3Î¸ âˆ’ cos 2Î¸ = 0

=> 2cos 2Î¸ cos Î¸ âˆ’ cos 2Î¸ = 0

=> cos 2Î¸ (2cos Î¸ âˆ’ 1) = 0

=> cos 2Î¸ = 0 or cos Î¸ = 1/2

=> 2Î¸ = (2n+1)Ï€/2 or cos Î¸ = cos Ï€/3

=> Î¸ = (2n+1)Ï€/4 or Î¸ = 2nÏ€ Â± (Ï€/3), n âˆˆ  Z

### (iii) sin Î¸ + sin 5Î¸ = sin 3Î¸

Solution:

We are given,

=> sin Î¸ + sin 5Î¸ = sin 3Î¸

=> 2sin 3Î¸ cos 2Î¸ âˆ’ sin 3Î¸ = 0

=> sin 3Î¸ (2cos 2Î¸ âˆ’ 1) = 0

=> sin 3Î¸ = 0 or cos 2Î¸ = 1/2

=> sin 3Î¸ = 0 or cos 2Î¸ = cos Ï€/3

=> 3Î¸ = nÏ€ or 2Î¸ = 2nÏ€ Â± (Ï€/3)

=> Î¸ = nÏ€/3 or Î¸ = nÏ€ Â± (Ï€/6), n âˆˆ  Z

### (iv) cos Î¸ cos 2Î¸ cos 3Î¸ = 1/4

Solution:

We are given,

=> cos Î¸ cos 2Î¸ cos 3Î¸ = 1/4

=> 2cos Î¸ cos 3Î¸ cos 2Î¸ = 1/2

=> (cos 4Î¸ + cos 2Î¸) cos 2Î¸ = 1/2

=> (2cos2 2Î¸ âˆ’ 1 + cos 2Î¸) cos 2Î¸ = 1/2

=> 2cos3 2Î¸ âˆ’ cos 2Î¸ + cos2 2Î¸ = 1/2

=> 4cos3 2Î¸ âˆ’ 2cos 2Î¸ + 2cos2 2Î¸ âˆ’ 1 = 0

=> 2cos2 2Î¸ (2cos 2Î¸ + 1) âˆ’ (2cos 2Î¸ + 1) = 0

=> (2cos2 2Î¸ âˆ’ 1) (2cos 2Î¸ + 1) = 0

=> 2cos2 2Î¸ âˆ’ 1 = 0 or 2cos 2Î¸ + 1 = 0

=> cos 4Î¸ = 0 or cos 2Î¸ = âˆ’1/2

=> 4Î¸ = (2n+1)Ï€/2 or cos 2Î¸ = cos 2Ï€/3

=> Î¸ = (2n+1)Ï€/8 or 2Î¸ = 2nÏ€ Â± (2Ï€/3)

=> Î¸ = (2n+1)Ï€/8 or 2Î¸ = nÏ€ Â± (Ï€/3), n âˆˆ  Z

### (v) cos Î¸ + sin Î¸ = cos 2Î¸ + sin 2Î¸

Solution:

We are given,

=> cos Î¸ + sin Î¸ = cos 2Î¸ + sin 2Î¸

=> cos Î¸ âˆ’ cos 2Î¸ = sin 2Î¸ âˆ’ sin Î¸

=> 2 sin 3Î¸/2 sin Î¸/2 = 2 cos 3Î¸/2 sin Î¸/2

=> 2 sin Î¸/2 (sin 3Î¸/2 âˆ’ cos 3Î¸/2) = 0

=> sin Î¸/2 = 0 or  (sin 3Î¸/2 âˆ’ cos 3Î¸/2) = 0

=> Î¸/2 = nÏ€ or tan 3Î¸/2 = 1

=> Î¸ = 2nÏ€ or tan 3Î¸/2 = tan Ï€/4

=> Î¸ = 2nÏ€ or 3Î¸/2 = nÏ€ Â± Ï€/4

=> Î¸ = 2nÏ€ or Î¸ = 2nÏ€/3 Â± Ï€/6, n âˆˆ  Z

### (vi) sin Î¸ + sin 2Î¸ + sin 3Î¸ = 0

Solution:

We are given,

=> (sin Î¸ + sin 3Î¸) + sin 2Î¸ = 0

=> 2sin 2Î¸ cos Î¸ + sin 2Î¸ = 0

=> sin 2Î¸ (2cos Î¸ + 1) = 0

=> sin 2Î¸ = 0 or 2cos Î¸ + 1 = 0

=> 2Î¸ = nÏ€ or cos Î¸ = âˆ’1/2

=> Î¸ = nÏ€/2 or cos Î¸ = cos 2Ï€/3

=> Î¸ = nÏ€/2 or Î¸ = 2nÏ€ Â± (2Ï€/3), n âˆˆ  Z

### (vii) sin x + sin 2x + sin 3x +  sin 4x = 0

Solution:

We are given,

=> sin x + sin 2x + sin 3x +  sin 4x = 0

=> (sin x + sin 3x) + (sin 2x + sin 4x) = 0

=> 2 sin 2x cos x + 2 sin 3x cos x = 0

=> 2cos x (sin 2x + sin 3x) = 0

=> (2cos x) (2sin 5x/2) (cos x/2) = 0

=> cos x = 0 or sin 5x/2 = 0 or cos x/2 = 0

=> x = (2n+1)Ï€/2 or 5x/2 = nÏ€ or cos x/2 = 0

=> x = (2n+1)Ï€/2 or x = 2nÏ€/5 or x/2 = (2n+1)Ï€/2

=> x = (2n+1)Ï€/2 or x = 2nÏ€/5 or x = (2n+1)Ï€, n âˆˆ  Z

### (viii) sin 3Î¸ âˆ’ sin Î¸ = 4 cos2 Î¸ âˆ’ 2

Solution:

We are given,

=> sin 3Î¸ âˆ’ sin Î¸ = 4 cos2 Î¸ âˆ’ 2

=> 2 cos 2Î¸ sin Î¸ = 2 (2cos2 Î¸ âˆ’ 1)

=> 2 cos 2Î¸ sin Î¸ = 2 cos 2Î¸

=> 2 cos 2Î¸ (sin Î¸ âˆ’ 1) = 0

=> cos 2Î¸ = 0 or (sin Î¸ âˆ’ 1) = 0

=> 2Î¸ = (2n+1)Ï€/2 or sin Î¸ = 1

=> Î¸ = (2n+1)Ï€/4 or sin Î¸ = sin Ï€/2

=> Î¸ = (2n+1)Ï€/4 or Î¸ = nÏ€ + (âˆ’1)n (Ï€/2), n âˆˆ  Z

### (ix) sin 2xâˆ’ sin 4x + sin 6x = 0

Solution:

We are given,

=> sin 2xâˆ’ sin 4x + sin 6x = 0

=> 2 sin 4x cos 2x âˆ’ sin 4x = 0

=> 2 sin 4x (cos 2x âˆ’ 1) = 0

=> sin 4x = 0 or cos 2x = 1/2

=> 4x = nÏ€ or cos 2x = cos Ï€/3

=> x = nÏ€/4 or 2x = 2nÏ€ Â± (Ï€/3)

=> x = nÏ€/4 or x = nÏ€ Â± (Ï€/6), n âˆˆ  Z

### (i) tan x + tan 2x + tan 3x = 0

Solution:

We are given,

=> tan x + tan 2x + tan 3x = 0

=> (tan x + tan 2x) + tan (2x+x) = 0

=> (tan x + tan 2x) + (tan x + tan 2x)/(1 âˆ’ tan x tan 2x) = 0

=> (tan x + tan 2x) [1 + 1/(1 âˆ’ tan x tan 2x)] = 0

=> (tan x + tan 2x) (2 âˆ’ tan x tan 2x) = 0

=> (tan x + tan 2x) = 0 or (2 âˆ’ tan x tan 2x) = 0

=> tan x = âˆ’tan 2x or tan x tan 2x = 2

=> tan x = tan (âˆ’2x) or tan x [2tan x/(1âˆ’tan2 x)] = 2

=> x = nÏ€ âˆ’ 2x or 2tan2 x = 2 âˆ’ tan2 x

=> 3x = nÏ€ or 4tan2 x = 2

=> x = nÏ€/3 or tan2 x = 1/2

=> x = nÏ€/3 or tan x = 1/âˆš2 or tan x = âˆ’1/âˆš2

=> x = nÏ€/3 or tan x = tan Ï€/4 or tan x = tan (âˆ’Ï€/4)

=> x = nÏ€/3 or x = nÏ€ + Ï€/4  or x = nÏ€ âˆ’ Ï€/4

=> x = nÏ€/3 or x = nÏ€ Â± Ï€/4, n âˆˆ  Z

### (ii) tan Î¸ + tan 2Î¸ = tan 3Î¸

Solution:

We are given,

=> tan Î¸ + tan 2Î¸ = tan 3Î¸

=> (tan Î¸ + tan 2Î¸) âˆ’ tan (2Î¸+Î¸) = 0

=> (tan Î¸ + tan 2Î¸) âˆ’ (tan Î¸ + tan 2Î¸)/(1 âˆ’ tan Î¸ tan 2Î¸) = 0

=> (tan Î¸ + tan 2Î¸) [1 âˆ’ 1/(1 âˆ’ tan Î¸ tan 2Î¸)] = 0

=> (tan Î¸ + tan 2Î¸) (âˆ’tan Î¸ tan 2Î¸) = 0

=> (tan Î¸ + tan 2Î¸) = 0 or âˆ’tan Î¸ = 0 or tan 2Î¸ = 0

=> tan Î¸ = tan (âˆ’2Î¸) or tan Î¸ = 0 or tan 2Î¸ = 0

=> Î¸ = nÏ€ âˆ’ 2Î¸ or Î¸ = nÏ€ or 2Î¸ = nÏ€

=> Î¸ = nÏ€/3 or Î¸ = nÏ€ or Î¸ = nÏ€/2, n âˆˆ  Z

### (iii) tan 3Î¸ + tan Î¸ = 2tan 2Î¸

Solution:

We are given,

=> tan 3Î¸ + tan Î¸ = 2tan 2Î¸

=> tan 3Î¸ âˆ’ tan 2Î¸ = tan 2Î¸ âˆ’ tan Î¸

=> 2 sin2 Î¸ sin 2Î¸ = 0

=> sin Î¸ = 0 or sin 2Î¸ = 0

=> Î¸ = nÏ€ or Î¸ = nÏ€/2, n âˆˆ  Z

### (i) sin Î¸ + cos Î¸ = âˆš2

Solution:

We are given,

=> sin Î¸ + cos Î¸ = âˆš2

=> (1/âˆš2) sin Î¸ + (1/âˆš2) cos Î¸ = 1

=> sin Ï€/4 sin Î¸ + cos Ï€/4 cos Î¸ = 1

=> cos (Î¸ âˆ’ Ï€/4) = cos 0

=> Î¸ âˆ’ Ï€/4 = 2nÏ€

=> Î¸ = 2nÏ€ + Ï€/4

=> Î¸ = (8n+1)Ï€/4, n âˆˆ  Z

### (ii) âˆš3 cos Î¸ + sin Î¸ = 1

Solution:

We are given,

=> âˆš3 cos Î¸ + sin Î¸ = 1

=> (âˆš3/2) cos Î¸ + (1/2) sin Î¸ = 1/2

=> cos Ï€/6 cos Î¸ + sin Ï€/6 sin Î¸ = 1/2

=> cos (Î¸ âˆ’ Ï€/6) = cos Ï€/3

=> Î¸ âˆ’ Ï€/6 = 2nÏ€ Â± Ï€/3

=> Î¸ = 2nÏ€ Â± Ï€/3 + Ï€/6

=> Î¸ = 2nÏ€ + Ï€/3 âˆ’ Ï€/6 or Î¸ = 2nÏ€ âˆ’ Ï€/3 + Ï€/6

=> Î¸ = 2nÏ€ + Ï€/2  or Î¸ = 2nÏ€ âˆ’ Ï€/6

=> Î¸ = (4n+1)Ï€/2  or Î¸ = (12nâˆ’1)Ï€/6, n âˆˆ  Z

### (iii) sin Î¸ + cos Î¸ = 1

Solution:

We are given,

=> sin Î¸ + cos Î¸ = 1

=> (1/âˆš2) cos Î¸ + (1/âˆš2) sin Î¸ = 1/âˆš2

=> cos Ï€/4 cos Î¸ + sin Ï€/4 sin Î¸ = 1/âˆš2

=> cos (Î¸ âˆ’ Ï€/4) = cos Ï€/4

=> Î¸ âˆ’ Ï€/4 = 2nÏ€ Â± Ï€/4

=> Î¸ = 2nÏ€ Â± Ï€/4 + Ï€/4

=> Î¸ = 2nÏ€ + Ï€/2 or Î¸ = 2nÏ€, n âˆˆ  Z

### (iv) cosec Î¸ = 1 + cot Î¸

Solution:

We are given,

=> cosec Î¸ = 1 + cot Î¸

=> 1/sin Î¸ = 1 + cos Î¸/sin Î¸

=> sin Î¸ + cos Î¸ = 1

=> (1/âˆš2) cos Î¸ + (1/âˆš2) sin Î¸ = 1/âˆš2

=> cos Ï€/4 cos Î¸ + sin Ï€/4 sin Î¸ = 1/âˆš2

=> cos (Î¸ âˆ’ Ï€/4) = cos Ï€/4

=> Î¸ âˆ’ Ï€/4 = 2nÏ€ Â± Ï€/4

=> Î¸ = 2nÏ€ Â± Ï€/4 + Ï€/4

=> Î¸ = 2nÏ€ + Ï€/2 or Î¸ = 2nÏ€, n âˆˆ  Z

### (v) (âˆš3 âˆ’ 1) cos Î¸ + (âˆš3 + 1) sin Î¸ = 2

Solution:

We are given,

=> (âˆš3 âˆ’ 1) cos Î¸ + (âˆš3 + 1) sin Î¸ = 2

=> (âˆš3 âˆ’ 1) cos Î¸/2âˆš2 + (âˆš3 + 1) sin Î¸/2âˆš2 = 2

=> sin (Î¸ + tan-1 (âˆš3 âˆ’ 1)/(âˆš3 + 1)) = sin Ï€/4

=> Î¸ = 2nÏ€ + Ï€/3 or Î¸ = 2nÏ€ âˆ’ Ï€/6, n âˆˆ  Z

### (i) cot x + tan x = 2

Solution:

We are given,

=> cot x + tan x = 2

=> cos x/sin x + sin x/cos x = 2

=> (cos2 x + sin2 x)/sin x cos x = 2

=> 2 sin x cos x = 1

=> sin 2x = 1

=> sin 2x = sin Ï€/2

=> 2x = (2n+1)Ï€/2

=> x = (2n+1)Ï€/4, n âˆˆ  Z

### (ii) 2 sin2 Î¸ = 3 cos Î¸, 0 â‰¤ Î¸ â‰¤ 2Ï€

Solution:

We are given,

=> 2 sin2 Î¸ = 3 cos Î¸

=> 2 (1 âˆ’ cos2 Î¸) âˆ’ 3 cos Î¸ = 0

=> 2 cos2 Î¸ + 3 cos Î¸ âˆ’ 2 = 0

=> 2 cos2 Î¸ + 4 cos Î¸ âˆ’ cos Î¸ âˆ’ 2 = 0

=> 2 cos Î¸ (cos + 2) âˆ’ (cos Î¸ + 2) = 0

=> (2 cos Î¸ âˆ’ 1) (cos Î¸ + 2) = 0

=> cos Î¸ = 1/2 or cos Î¸ = âˆ’2

Ignoring cos Î¸ =âˆ’2  as âˆ’1 â‰¤ cos Î¸ â‰¤ 1. So, we have, cos Î¸ = 1/2.

=> cos Î¸ = cos Ï€/3

=> Î¸ = 2nÏ€ Â± Ï€/3, n âˆˆ  Z

### (iii) sec x cos 5x +1 = 0, 0 â‰¤ x â‰¤ Ï€/2

Solution:

We are given,

=> sec x cos 5x +1 = 0

=> (cos5x + cos x)/cos x = 0

=> 2 cos 3x cos 2x = 0

=> cos 3x = 0 or cos 2x = 0

=> 3x = (2n+1)Ï€/2 or 2x = (2n+1)Ï€/2

=> x = (2n+1)Ï€/6 or x = (2n+1)Ï€/4, n âˆˆ  Z

### (iv) 5 cos2 Î¸ + 7 sin2 Î¸ âˆ’ 6 = 0

Solution:

We are given,

=> 5 cos2 Î¸ + 7 sin2 Î¸ âˆ’ 6 = 0

=> 5 (1 âˆ’ sin2 Î¸) + 7 sin2 Î¸ âˆ’ 6 = 0

=> 2 sin2 Î¸+ 5 âˆ’ 6 = 0

=> 2 sin2 Î¸ = 1

=> sin Î¸ = Â±(1/âˆš2)

=> Î¸ = nÏ€ Â± Ï€/4, n âˆˆ  Z

### (v) sin x âˆ’ 3 sin 2x + sin 3x = cos x âˆ’ 3 cos 2x + cos 3x

Solution:

We are given,

=> sin x âˆ’ 3 sin 2x + sin 3x = cos x âˆ’ 3 cos 2x + cos 3x

=> (sin x + sin 3x) âˆ’ 3 sin 2x âˆ’ (cos x + cos 3x) + 3 cos 2x = 0

=> 2 sin 2x cos x âˆ’ 3 sin 2x âˆ’ 2 cos 2x cos x + 3 cos 2x = 0

=> sin 2x (2 cos x âˆ’ 3) âˆ’ cos 2x (2 cos x âˆ’ 3) = 0

=> (sin 2x âˆ’ cos 2x) (2 cos x âˆ’ 3) = 0

=> sin 2x = cos 2x or cos x = 3/2

Ignoring cos x = 3/2 as âˆ’1 â‰¤ cos x â‰¤ 1. So, we have, sin 2x = cos 2x.

=> tan 2x = 1

=> tan 2x = tan Ï€/4

=> 2x = nÏ€ + Ï€/4

=> x = nÏ€/2 + Ï€/8, n âˆˆ  Z

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