If sin 3y = cos 2y and 0^o $\leq$ 90^o, find the value of y

18^o

36^o

54^o

90^o

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Correct answer is A

sin 3y = cos 2y, but sin $\theta$ = cos(90 - $\theta$ )

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = $\frac{90}{5}$

y = 18^o

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If sin 3y = cos 2y and 0o ≤\leq 90o, find the value of y

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If sin 3y = cos 2y and 0^o $\leq$ 90^o, find the value of y