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

• Last Updated : 28 Apr, 2021

### (i) sin θ = 1/2

Solution:

We are given,

=> sin θ = 1/2

=> sin θ = sin π/6

We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a; n ∈ Z.

=> θ = nπ + (−1)n (π/6) ; n ∈ Z

### (ii) cos θ = −√3/2

Solution:

We are given,

=> cos θ = −√3/2

=> cos θ = cos (π + π/6)

=> cos θ = cos (7π/6)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈  Z.

=> θ = 2nπ ± (7π/6) ; n ∈  Z

### (iii) cosec θ = −√2

Solution:

We are given,

=> cosec θ = −√2

=> 1/sin θ = −√2

=> sin θ = −1/√2

=> sin θ = −sin (π/4)

As sin (−θ) = − sin θ, we have,

=> sin θ = sin (−π/4)

We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a ; n ∈  Z.

=> θ = nπ + (−1)n+1 (π/4) ; n ∈  Z

### (iv) sec θ = √2

Solution:

We are given,

=> sec θ = √2

=> 1/sec θ = √2

=> cos θ = 1/√2

=> cos θ = cos π/4

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈  Z.

=> θ = 2nπ ± (π/4) ; n ∈  Z

### (v) tan θ = −1/√3

Solution:

We are given,

=> tan θ = −1/√3

=> tan θ = −tan (π/6)

As tan (−θ) = − tan θ, we have,

=> tan θ = tan (−π/6)

We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈  Z.

=> θ = nπ − (π/6) ; n ∈  Z

### (vi) √3 sec θ = 2

Solution:

We are given,

=> √3 sec θ = 2

=> √3 (1/cos θ) = 2

=> cos θ = √3/2

=> cos θ = cos (π/6)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a; n ∈ Z.

=> θ = 2nπ ± (π/6) ; n ∈ Z

### (i) sin 2θ = √3/2

Solution:

We are given,

=> sin 2θ = √3/2

=> sin 2θ = sin (π/3)

We know if sin θ = sin a, the general solution is given by, θ = nπ + (−1)n a; n ∈ Z.

=> 2θ = nπ + (−1)n (π/3)

=> θ = nπ/2 + (−1)n (π/6) ; n ∈ Z

### (ii) cos 3θ = 1/2

Solution:

We are given,

=> cos 3θ = 1/2

=> cos 3θ = cos (π/3)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a; n ∈ Z.

=> 3θ = 2nπ ± (π/3)

=> θ = 2nπ/3 ± (π/9) ; n ∈ Z

### (iii) sin 9θ = sin θ

Solution:

We are given,

=> sin 9θ = sin θ

=> sin 9θ – sin θ = 0

=> 2 cos 5θ sin 4θ = 0

=> cos 5θ = 0 or sin 4θ = 0

We know, if cos θ = 0, then θ = (2n+1)π/2 and if sin θ = 0, then θ = nπ where n ∈  Z.

=> 5θ = (2n+1)π/2 or 4θ = nπ

=> θ = (2n+1)π/10 or θ = nπ/4, n ∈  Z

### (iv) sin 2θ = cos 3θ

Solution:

We are given,

=> sin 2θ = cos 3θ

=> cos 3θ = sin 2θ

=> cos 3θ = cos (π/2 − 2θ)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈  Z.

=> 3θ = 2nπ ± (π/2 − 2θ)

=> 3θ = 2nπ + π/2 − 2θ or 3θ = 2nπ − π/2 + 2θ

=> 5θ = 2nπ + π/2 or θ = 2nπ − π/2

=> 5θ = (4n+1) (π/2) or θ = (4n−1) (π/2)

=> θ = (4n+1)π/10 or θ = (4n−1)π/2, n ∈  Z

### (v) tan θ + cot 2θ = 0

Solution:

We are given,

=> tan θ + cot 2θ = 0

=> cot 2θ = − tan θ

=> tan 2θ = − cot θ

=> tan 2θ = − tan (π/2 − θ)

As tan (−θ) = − tan θ, we have,

=> tan 2θ = tan (θ − π/2)

We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈  Z.

=> 2θ = nπ + θ − π/2

=> θ = nπ − π/2 ; n ∈  Z

### (vi) tan 3θ = cot θ

Solution:

We are given,

=> tan 3θ = cot θ

=> tan 3θ = tan (π/2 − θ)

We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈  Z.

=> 3θ = nπ + π/2 − θ

=> 4θ = nπ + π/2

=> θ = nπ/4 + π/8 ; n ∈  Z

### (vii) tan 2θ tan θ = 1

Solution:

We are given,

=> tan 2θ tan θ = 1

=> tan 2θ = cot θ

=> tan 2θ = tan (π/2 − θ)

We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈  Z.

=> 2θ = nπ + π/2 − θ

=> 3θ = nπ + π/2

=> θ = nπ/3 + π/6 ; n ∈  Z

### (viii) tan mθ + cot nθ = 0

Solution:

We are given,

=> tan mθ + cot nθ = 0

=> sin mθ/cos mθ + cos nθ/sin nθ = 0

=> sin mθ sin nθ + cos nθ cos mθ = 0

=> cos (m−n)θ = 0

We know, if cos θ = 0, then θ = (2k+1)π/2 where k ∈  Z.

=> (m−n)θ = (2k+1)π/2

=> θ = (2k+1)π/2(m−n) ; n ∈  Z

### (ix) tan pθ = cot qθ

Solution:

We are given,

=> tan pθ = cot qθ

=> tan pθ = tan (π/2 − qθ)

We know if tan θ = tan a, the general solution is given by, θ = nπ + a ; n ∈  Z.

=> pθ = nπ + π/2 − qθ

=> (p + q)θ = (2n+1)π/2

=> θ = (2n+1)π/2(p + q) ; n ∈  Z

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

Solution:

We are given,

=> sin 2x + cos x = 0

=> cosx = − sin 2x

=> cos x = cos (π/2 + 2x)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈  Z.

=> x = 2nπ ± (π/2 + 2x)

=> x = 2nπ + π/2 + 2x or x =  2nπ − π/2 − 2x

=> x = −(4n+1)π/2 or 3x = (4n−1)π/2

=> x = −(4n+1)π/2 or 3x = (4n−1)π/2

=> x = (4m−1)π/2, where m = −n or x = (4n−1)π/6, n ∈  Z

### (xi) sin θ = tan θ

Solution:

We are given,

=> sin θ = tan θ

=> sin θ = sin θ/cos θ

=> sin θ (cos θ − 1) = 0

=> sin θ = 0 or cos θ = 1

=> sin θ = 0 or cos θ = cos 0

We know, if cos θ = 0, then θ = (2n+1)π/2 and if sin θ = 0, then θ = nπ where n ∈  Z.

=> θ = nπ or θ = 2nπ, n ∈  Z

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

Solution:

We are given,

=> sin 3x + cos 2x = 0

=> cos 2x = −sin 3x

=>cos 2x = cos (π/2 + 3x)

We know if cos θ = cos a, the general solution is given by, θ = 2nπ ± a ; n ∈  Z.

=> 2x = 2nπ ± (π/2 + 3x)

=> 2x = 2nπ + π/2 + 3x or 2x = 2nπ − π/2 − 3x

=> −x = 2nπ + π/2 or 5x = 2nπ − π/2

=> x = 2mπ − π/2, where m = −n or x = (2nπ − π/2)/5

=> x = (4m−1)π/2, where m = −n or x = (4n−1)π/10, n ∈  Z

### (i) sin2 θ − cos θ = 1/4

Solution:

We have,

=> sin2 θ − cos θ = 1/4

=> 1 − cos2 θ − cos θ = 1/4

=> cos2 θ + cos θ − 3/4 = 0

=> 4cos2 θ + 4cos θ − 3 = 0

=> 4cos2 θ + 6cos θ − 2cos θ − 3 = 0

=> 2cos θ (2cos + 3) − (2cos θ + 3) = 0

=> (2cos θ − 1) (2cos θ + 3) = 0

=> cos θ = 1/2 or cos θ = −3/2

Ignoring cos θ = −3/2 as −1 ≤ cosθ ≤ 1. So, we have, cos θ = 1/2.

=> cos θ = cos π/3

=> θ = 2nπ ± π/3 ; n ∈  Z

### (ii) 2cos2 θ − 5cos θ + 2 = 0

Solution:

We are given,

=> 2cos2 θ − 5cos θ + 2 = 0

=> 2cos2 θ − 4cos θ − cos θ + 2 = 0

=> 2cos θ (cos θ − 2) − (cos θ − 2) = 0

=> (2cos θ − 1) (cos θ − 2) = 0

=> cos θ = 1/2 or cos θ = 2

Ignoring cos θ = 2 as −1 ≤ cosθ ≤ 1. So, we have, cos θ = 1/2.

=> cos θ = cos π/3

=> θ = 2nπ ± π/3 ; n ∈  Z

### (iii) 2sin2 x + √3cos x + 1 = 0

Solution:

We are given,

=> 2sin2 x + √3cos x + 1 = 0

=> 2 (1 − cos2 x) + √3cos x + 1 = 0

=> 2cos2 x − √3cos x − 3 = 0

=> 2cos2 x − 2√3cos x + √3cos x − 3 = 0

=> 2cosx (cos x − √3) + √3(cos x − √3) = 0

=> (2cosx + √3) (cos x − √3) = 0

=> x = −√3/2 or x = √3

Ignoring cos x = √3 as −1 ≤ cosθ ≤ 1. So, we have, cos x = −√3/2.

=> cos x = − cos π/6

=> cos x = cos (π − π/6)

=> cos x = cos (5π/6)

=> x = 2nπ ± 5π/6 ; n ∈  Z

### (iv) 4sin2 θ − 8cos θ + 1 = 0

Solution:

We are given,

=> 4sin2 θ − 8cos θ + 1 = 0

=> 4 (1 − cos2 θ) − 8cos θ + 1 = 0

=> 4 cos2 θ + 8 cos θ − 5 = 0

=> 4 cos2 θ + 10 cos θ − 2 cos θ − 5 = 0

=> 2cos θ (2cos θ +5) − (2cos θ +5) = 0

=> (2cos θ − 1) (2cos θ +5) = 0

=> cos θ = 1/2 or cos θ = −5/2

Ignoring cos x = −5/2 as −1 ≤ cosθ ≤ 1. So, we have, cos x = 1/2.

=> cos x = cos π/3

=> x = 2nπ ± π/3 ; n ∈  Z

### (v) tan2 x + (1 − √3)tan x − √3 = 0

Solution:

We are given,

=> tan2 x + (1 − √3)tan x − √3 = 0

=> tan2 x + tan x − √3tan x − √3 = 0

=> tan x (tan x + 1) − √3 (tan x + 1) = 0

=> (tan x − √3) (tan x + 1) = 0

=> tan x = √3 or tan x = −1

=> tan x = tan π/3 or tan x = −tan π/4

=> tan x = tan π/3 or tan x = tan (−π/4)

=> x = nπ + π/3 or x = nπ − π/4, n ∈  Z

### (vi) 3 cos2 θ − 2√3 sin θ cos θ − 3 sin2 θ = 0

Solution:

We have,

=> 3 cos2 θ − 2√3 sin θ cos θ − 3 sin2 θ = 0

=> √3 cos2 θ − 2 sin θ cos θ − √3 sin2 θ = 0

=> √3 cos2 θ + sin θ cos θ − 3 sin θ cos θ − √3 sin2 θ = 0

=> cos θ (√3cos θ + sin θ) − √3sin θ (√3cos θ + sin θ) = 0

=> (√3cos θ + sin θ) (cos θ − √3sin θ) = 0

=> tan θ = −√3 or tan θ = 1/√3

=> tan θ = − tan π/3 or tan θ = tan π/6

=> tan θ = tan (−π/3) or tan θ = tan π/6

=> θ = nπ − π/3 or θ = nπ + π/6, n ∈  Z

### (vii) cos 4θ = cos 2θ

Solution:

We have,

=> cos 4θ = cos 2θ

=>  cos 4θ − cos 2θ = 0

=> 2 sin 3θ sin θ = 0

=> sin 3θ = 0 or sin θ = 0

=> 3θ = nπ or θ = nπ

=> θ = nπ/3 or θ = nπ, n ∈  Z

### (i) cos θ + cos 2θ + cos 3θ = 0

Solution:

We are given,

=> cos 2θ + (cos θ + cos 3θ) = 0

=> cos 2θ + 2cos 2θ cos θ = 0

=> cos 2θ (1 + 2cos θ) = 0

=> cos 2θ = 0 or cos θ = −1/2

=> 2θ = (2n+1)π/2 or cos θ = cos (π − π/3)

=> θ = (2n+1)π/4 or θ = 2nπ ± (2π/3), n ∈  Z

### (ii) cos θ + cos 3θ − cos 2θ = 0

Solution:

We are given,

=> cos θ + cos 3θ − cos 2θ = 0

=> 2cos 2θ cos θ − cos 2θ = 0

=> cos 2θ (2cos θ − 1) = 0

=> cos 2θ = 0 or cos θ = 1/2

=> 2θ = (2n+1)π/2 or cos θ = cos π/3

=> θ = (2n+1)π/4 or θ = 2nπ ± (π/3), n ∈  Z

### (iii) sin θ + sin 5θ = sin 3θ

Solution:

We are given,

=> sin θ + sin 5θ = sin 3θ

=> 2sin 3θ cos 2θ − sin 3θ = 0

=> sin 3θ (2cos 2θ − 1) = 0

=> sin 3θ = 0 or cos 2θ = 1/2

=> sin 3θ = 0 or cos 2θ = cos π/3

=> 3θ = nπ or 2θ = 2nπ ± (π/3)

=> θ = nπ/3 or θ = nπ ± (π/6), n ∈  Z

### (iv) cos θ cos 2θ cos 3θ = 1/4

Solution:

We are given,

=> cos θ cos 2θ cos 3θ = 1/4

=> 2cos θ cos 3θ cos 2θ = 1/2

=> (cos 4θ + cos 2θ) cos 2θ = 1/2

=> (2cos2 2θ − 1 + cos 2θ) cos 2θ = 1/2

=> 2cos3 2θ − cos 2θ + cos2 2θ = 1/2

=> 4cos3 2θ − 2cos 2θ + 2cos2 2θ − 1 = 0

=> 2cos2 2θ (2cos 2θ + 1) − (2cos 2θ + 1) = 0

=> (2cos2 2θ − 1) (2cos 2θ + 1) = 0

=> 2cos2 2θ − 1 = 0 or 2cos 2θ + 1 = 0

=> cos 4θ = 0 or cos 2θ = −1/2

=> 4θ = (2n+1)π/2 or cos 2θ = cos 2π/3

=> θ = (2n+1)π/8 or 2θ = 2nπ ± (2π/3)

=> θ = (2n+1)π/8 or 2θ = nπ ± (π/3), n ∈  Z

### (v) cos θ + sin θ = cos 2θ + sin 2θ

Solution:

We are given,

=> cos θ + sin θ = cos 2θ + sin 2θ

=> cos θ − cos 2θ = sin 2θ − sin θ

=> 2 sin 3θ/2 sin θ/2 = 2 cos 3θ/2 sin θ/2

=> 2 sin θ/2 (sin 3θ/2 − cos 3θ/2) = 0

=> sin θ/2 = 0 or  (sin 3θ/2 − cos 3θ/2) = 0

=> θ/2 = nπ or tan 3θ/2 = 1

=> θ = 2nπ or tan 3θ/2 = tan π/4

=> θ = 2nπ or 3θ/2 = nπ ± π/4

=> θ = 2nπ or θ = 2nπ/3 ± π/6, n ∈  Z

### (vi) sin θ + sin 2θ + sin 3θ = 0

Solution:

We are given,

=> (sin θ + sin 3θ) + sin 2θ = 0

=> 2sin 2θ cos θ + sin 2θ = 0

=> sin 2θ (2cos θ + 1) = 0

=> sin 2θ = 0 or 2cos θ + 1 = 0

=> 2θ = nπ or cos θ = −1/2

=> θ = nπ/2 or cos θ = cos 2π/3

=> θ = nπ/2 or θ = 2nπ ± (2π/3), n ∈  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)π/2 or 5x/2 = nπ or cos x/2 = 0

=> x = (2n+1)π/2 or x = 2nπ/5 or x/2 = (2n+1)π/2

=> x = (2n+1)π/2 or x = 2nπ/5 or x = (2n+1)π, n ∈  Z

### (viii) sin 3θ − sin θ = 4 cos2 θ − 2

Solution:

We are given,

=> sin 3θ − sin θ = 4 cos2 θ − 2

=> 2 cos 2θ sin θ = 2 (2cos2 θ − 1)

=> 2 cos 2θ sin θ = 2 cos 2θ

=> 2 cos 2θ (sin θ − 1) = 0

=> cos 2θ = 0 or (sin θ − 1) = 0

=> 2θ = (2n+1)π/2 or sin θ = 1

=> θ = (2n+1)π/4 or sin θ = sin π/2

=> θ = (2n+1)π/4 or θ = nπ + (−1)n (π/2), n ∈  Z

### (ix) sin 2x− sin 4x + sin 6x = 0

Solution:

We are given,

=> sin 2x− sin 4x + sin 6x = 0

=> 2 sin 4x cos 2x − sin 4x = 0

=> 2 sin 4x (cos 2x − 1) = 0

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

=> 4x = nπ or cos 2x = cos π/3

=> x = nπ/4 or 2x = 2nπ ± (π/3)

=> x = nπ/4 or x = nπ ± (π/6), n ∈  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 − tan x tan 2x) = 0

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

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

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

=> tan x = −tan 2x or tan x tan 2x = 2

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

=> x = nπ − 2x or 2tan2 x = 2 − tan2 x

=> 3x = nπ or 4tan2 x = 2

=> x = nπ/3 or tan2 x = 1/2

=> x = nπ/3 or tan x = 1/√2 or tan x = −1/√2

=> x = nπ/3 or tan x = tan π/4 or tan x = tan (−π/4)

=> x = nπ/3 or x = nπ + π/4  or x = nπ − π/4

=> x = nπ/3 or x = nπ ± π/4, n ∈  Z

### (ii) tan θ + tan 2θ = tan 3θ

Solution:

We are given,

=> tan θ + tan 2θ = tan 3θ

=> (tan θ + tan 2θ) − tan (2θ+θ) = 0

=> (tan θ + tan 2θ) − (tan θ + tan 2θ)/(1 − tan θ tan 2θ) = 0

=> (tan θ + tan 2θ) [1 − 1/(1 − tan θ tan 2θ)] = 0

=> (tan θ + tan 2θ) (−tan θ tan 2θ) = 0

=> (tan θ + tan 2θ) = 0 or −tan θ = 0 or tan 2θ = 0

=> tan θ = tan (−2θ) or tan θ = 0 or tan 2θ = 0

=> θ = nπ − 2θ or θ = nπ or 2θ = nπ

=> θ = nπ/3 or θ = nπ or θ = nπ/2, n ∈  Z

### (iii) tan 3θ + tan θ = 2tan 2θ

Solution:

We are given,

=> tan 3θ + tan θ = 2tan 2θ

=> tan 3θ − tan 2θ = tan 2θ − tan θ

=> 2 sin2 θ sin 2θ = 0

=> sin θ = 0 or sin 2θ = 0

=> θ = nπ or θ = nπ/2, n ∈  Z

### (i) sin θ + cos θ = √2

Solution:

We are given,

=> sin θ + cos θ = √2

=> (1/√2) sin θ + (1/√2) cos θ = 1

=> sin π/4 sin θ + cos π/4 cos θ = 1

=> cos (θ − π/4) = cos 0

=> θ − π/4 = 2nπ

=> θ = 2nπ + π/4

=> θ = (8n+1)π/4, n ∈  Z

### (ii) √3 cos θ + sin θ = 1

Solution:

We are given,

=> √3 cos θ + sin θ = 1

=> (√3/2) cos θ + (1/2) sin θ = 1/2

=> cos π/6 cos θ + sin π/6 sin θ = 1/2

=> cos (θ − π/6) = cos π/3

=> θ − π/6 = 2nπ ± π/3

=> θ = 2nπ ± π/3 + π/6

=> θ = 2nπ + π/3 − π/6 or θ = 2nπ − π/3 + π/6

=> θ = 2nπ + π/2  or θ = 2nπ − π/6

=> θ = (4n+1)π/2  or θ = (12n−1)π/6, n ∈  Z

### (iii) sin θ + cos θ = 1

Solution:

We are given,

=> sin θ + cos θ = 1

=> (1/√2) cos θ + (1/√2) sin θ = 1/√2

=> cos π/4 cos θ + sin π/4 sin θ = 1/√2

=> cos (θ − π/4) = cos π/4

=> θ − π/4 = 2nπ ± π/4

=> θ = 2nπ ± π/4 + π/4

=> θ = 2nπ + π/2 or θ = 2nπ, n ∈  Z

### (iv) cosec θ = 1 + cot θ

Solution:

We are given,

=> cosec θ = 1 + cot θ

=> 1/sin θ = 1 + cos θ/sin θ

=> sin θ + cos θ = 1

=> (1/√2) cos θ + (1/√2) sin θ = 1/√2

=> cos π/4 cos θ + sin π/4 sin θ = 1/√2

=> cos (θ − π/4) = cos π/4

=> θ − π/4 = 2nπ ± π/4

=> θ = 2nπ ± π/4 + π/4

=> θ = 2nπ + π/2 or θ = 2nπ, n ∈  Z

### (v) (√3 − 1) cos θ + (√3 + 1) sin θ = 2

Solution:

We are given,

=> (√3 − 1) cos θ + (√3 + 1) sin θ = 2

=> (√3 − 1) cos θ/2√2 + (√3 + 1) sin θ/2√2 = 2

=> sin (θ + tan-1 (√3 − 1)/(√3 + 1)) = sin π/4

=> θ = 2nπ + π/3 or θ = 2nπ − π/6, n ∈  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 π/2

=> 2x = (2n+1)π/2

=> x = (2n+1)π/4, n ∈  Z

### (ii) 2 sin2 θ = 3 cos θ, 0 ≤ θ ≤ 2π

Solution:

We are given,

=> 2 sin2 θ = 3 cos θ

=> 2 (1 − cos2 θ) − 3 cos θ = 0

=> 2 cos2 θ + 3 cos θ − 2 = 0

=> 2 cos2 θ + 4 cos θ − cos θ − 2 = 0

=> 2 cos θ (cos + 2) − (cos θ + 2) = 0

=> (2 cos θ − 1) (cos θ + 2) = 0

=> cos θ = 1/2 or cos θ = −2

Ignoring cos θ =−2  as −1 ≤ cos θ ≤ 1. So, we have, cos θ = 1/2.

=> cos θ = cos π/3

=> θ = 2nπ ± π/3, n ∈  Z

### (iii) sec x cos 5x +1 = 0, 0 ≤ x ≤ π/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)π/2 or 2x = (2n+1)π/2

=> x = (2n+1)π/6 or x = (2n+1)π/4, n ∈  Z

### (iv) 5 cos2 θ + 7 sin2 θ − 6 = 0

Solution:

We are given,

=> 5 cos2 θ + 7 sin2 θ − 6 = 0

=> 5 (1 − sin2 θ) + 7 sin2 θ − 6 = 0

=> 2 sin2 θ+ 5 − 6 = 0

=> 2 sin2 θ = 1

=> sin θ = ±(1/√2)

=> θ = nπ ± π/4, n ∈  Z

### (v) sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x

Solution:

We are given,

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

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

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

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

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

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

Ignoring cos x = 3/2 as −1 ≤ cos x ≤ 1. So, we have, sin 2x = cos 2x.

=> tan 2x = 1

=> tan 2x = tan π/4

=> 2x = nπ + π/4

=> x = nπ/2 + π/8, n ∈  Z

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