Question

Solve for $\theta$ if $4co{s}^2\theta$ + 2$sin\theta$ – 4 = 0, where 0 $\le$ $\theta$ $\le$ 90.

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. both (a) and (b)
• D. cannot be determined

Answer 1

The correct option is C

both (a) and (b)

$4co{s}^2\theta$ + 2$sin\theta$ – 4 = 0

4[1 - $si{n}^2\theta$] + 2$sin\theta$ – 4 = 0

4 - 4$si{n}^2\theta$ + 2$sin\theta$ – 4 = 0

2$sin\theta$ [1 - 2$sin\theta$] = 0

$sin\theta$ = 0 or $sin\theta$ = $\frac{1}{2}$
So, $\theta =0^{\circ }or\theta ={30}^{\circ }$