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# How do you solve 4sinx-3sin2x=0 for 0<=x<=360?

How do you solve 4sinx-3sin2x=0 for 0<=x<=360?

ReportAnswer 1

$0 , {48}^{\circ} 19 , {180}^{\circ} , {311}^{\circ} 81 , {360}^{\circ}$

#### Explanation:

Since sin 2x = 2sin x.cos x, we get:
4sin x - 6sin x.cos x = 0
sin x(4 - 6cos x) = 0
Trig table and unit circle -->
a. sin x = 0 --> $x = 0$, $x = {180}^{\circ}$, and $x = {360}^{\circ}$
b. 4 - 6cos x = 0
cos x = 4/6 = 2/3 --> x = +- 48^@19
Arc (-48.19) and arc (360 - 48.19 = 311.81) are co-terminal
Answers for (0, 360)
$0 , {48}^{\circ} 19 , {180}^{\circ} , {311}^{\circ} 81 , {360}^{\circ}$

answered by: Nghi N.
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