How do you solve $sin (3 x) + {sin}^{2} (x) = 2$ $sin(3x) + sin^2 (x)= 2$ ?

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How do you solve $sin (3 x) + {sin}^{2} (x) = 2$ $sin(3x) + sin^2 (x)= 2$ ?

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Answer 1 · 2015-10-17T00:05:00

Solve sin 3x + sin^2 x = 2

Ans: $x = \frac{3 π}{2}$ $x = (3pi)/2$

Explanation:

$sin 3 x + {sin}^{2} x = 2$ $sin 3x + sin^2 x = 2$
Trig identity --> $sin (3 x) = 3 sin x - 4 {sin}^{3} x .$ $sin (3x) = 3sin x - 4sin^3 x.$ Call sin x = t, we get
$3 t - 4 t^{3} + t^{2} = 2$ $3t - 4t^3 + t^2 = 2$
$f (t) = 4 t^{3} - t^{2} - 3 t + 2 = 0$ $f(t) = 4t^3 - t^2 - 3t + 2 = 0$
Since (a - b + c - d = 0), one factor is (t + 1). By division, we get:
$f (t) = (t + 1) (4 t^{2} - 5 t + 2) = 0$ $f(t) = (t + 1)(4t^2 - 5t + 2) = 0$ . Solve this product.

a. t = sin x = - 1 --> $x = \frac{3 π}{2}$ $x = (3pi)/2$
b. $(4 t^{2} - 5 t + 2) = 0$ $(4t^2 - 5t + 2) = 0$
$D = b^{2} - 4 a c = 25 - 32 < 0 .$ $D = b^2 - 4ac = 25 - 32 < 0.$ There are no real roots.
Therefor, there is unique answer: $x = \frac{3 π}{2}$ $x = (3pi)/2$
Check.
$x = \frac{3 π}{2}$ $x = (3pi)/2$ --> $sin 3 x = sin (\frac{9 π}{2}) = sin (\frac{π}{2}) = 1$ $sin 3x = sin ((9pi)/2) = sin (pi/2) = 1$ --> $sin^2 x = 1.$
$sin 3x + sin^2 x = 1 + 1 = 2$ . OK

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