18 sin2 x − 35 sin x + 12 = 0

I need to find all the solutions in the interval
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1 Answer

18sin2(x)-35sin(x)+12=0=(2sin(x)-3)(9sin(x)-4).

Since sin(x) cannot exceed 1, sin(x)=4/9, x=sin-1(4/9)=26.3878° approx.

x can also be 180-26.3878=153.6122°. Other solutions are found by adding 360° to these values, or by subtracting 360°. No interval has been specified.

by Top Rated User (1.2m points)

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