Page 59 - mathsvol1ch1to3ans

59

1 π π π
when sin 2x = , 2x = (−1) n + nπ and hence x = (−1) n + n
2 6 12 2

2
(ii) 2 cos θ + 3 sin θ − 3 = 0
2
2(1 − sin θ) + 3 sin θ − 3 = 0
2
2 sin θ − 3 sin θ + 1 = 0
sin (θ) = 1, sin (θ) = 1
2
π

When sin (θ) = 1 ⇒ θ = nπ + (−1) n
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2
1
π
When sin (θ) = ⇒ θ = nπ + (−1) n
2 6
(iii) cos θ + cos 3θ = 2 cos 2θ
2
3
cos θ + 4 cos θ − 3 cos θ = 4 cos θ − 2
2
3
4 cos θ − 4 cos θ − 2 cos θ + 2 = 0
3
2
2 cos θ − 2 cos θ − cos θ + 1 = 0
2
2(cos θ)(cos θ − 1) − 1(cos θ − 1) = 0
2
(2 cos θ − 1)(cos θ − 1) = 0
1
cos θ = ±√ or cos θ = 1
2
π
Hence θ = 2nπ ± or θ = 2nπ
4
(iv) sin θ + sin 3θ + sin 5θ = 0

sin θ + sin 5θ + sin 3θ = 0

2 sin 3θ cos 2θ + sin 3θ = 0

sin 3θ(2 cos 2θ + 1) = 0

1
Hence sin 3θ = 0 or cos 2θ = −
2
π π
Solution is θ = nπ or 2θ = 2nπ ± ⇒ θ = nπ ± .
6 3

(v) sin 2θ − cos 2θ − sin θ + cos θ = 0

We have (sin 2θ − sin θ) − (cos 2θ − cos θ) = 0

3θ θ θ 3θ
Hence 2 sin cos + 2 sin sin = 0
2 2 2 2

3θ θ θ
sin cos + sin = 0
2 2 2

θ θ
cos + sin = 0
2 2