Page 60 - mathsvol1ch1to3ans

60

θ θ
Divide both sides by cos , cos 6= 0
2 2

θ θ
cos + sin
2 2 0
=

θ θ
cos cos
2 2

θ
tan + 1 = 0
2

θ
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tan + 1 − 1 = 0 − 1
2

θ
tan = −1
2
θ 3π
= + πn
2 4
3π
θ = + 2nπ
2
π

cos θ = 0 ⇒ θ = (2n + 1)
2
√
(vi) sin θ + cos θ = 2

1
Multiplying both sides by √
2
1 1
√ sin θ + √ cos θ = 1
2 2
π π
cos sin θ + sin cos θ = 1
4 4
π π

sin + θ = sin
4 2
π
θ =
4
π
Hence the solution is θ = nπ + (−1) n
√ 4
(vii) sin θ + 3 cos θ = 1

1
Multiplying both sides by
2
1 1 √ 1
sin θ + 3 cos θ =
2 2 2
π π π
cos sin θ + sin cos θ = sin
3 3 6
π π

sin + θ = sin
3 6
π
θ = −
3
π
Hence the solution is θ = nπ + (−1) n
3