What is the general solution to the equation $sin (3 x) - sin x = 1$ $sin(3x) - sinx = 1$ ?

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Answer 1 · 2016-12-04T14:36:10

Recall that $sin (3 x) = sin (2 x + x)$ $sin(3x) = sin(2x+ x)$ .

We use the sum formula $sin (A + B) = sin A cos B + sin B cos A$ $sin(A + B) = sinAcosB + sinBcosA$ to expand $sin (2 x + x)$ $sin(2x + x)$ .

$sin 2 x cos x + cos 2 x sin x - sin x = 1$ $sin2xcosx + cos2xsinx - sinx = 1$

$2 sin x cos x (cos x) + (2 {cos}^{2} x - 1) sin x - sin x = 1$ $2sinxcosx(cosx) +( 2cos^2x - 1)sinx - sinx = 1$

$2 sin x {cos}^{2} x + 2 {cos}^{2} x sin x - sin x - sin x = 1$ $2sinxcos^2x + 2cos^2xsinx - sinx - sinx = 1$

$4 sin x {cos}^{2} x - 2 sin x = 1$ $4sinxcos^2x - 2sinx = 1$

$2 sin x (2 {cos}^{2} x - 1) = 1$ $2sinx(2cos^2x- 1) = 1$

$2 {cos}^{2} x - 1 = \frac{1}{2 sin x}$ $2cos^2x - 1 = 1/(2sinx)$

$2 {cos}^{2} x = 1 + \frac{1}{2 sin x}$ $2cos^2x = 1+ 1/(2sinx)$

$2 (1 - {sin}^{2} x) = \frac{2 sin x + 1}{2 sin x}$ $2(1 - sin^2x) = (2sinx + 1)/(2sinx)$

$2 (1 - {sin}^{2} x) = \frac{2 (sin x + \frac{1}{2})}{2 sin x}$ $2 (1 - sin^2x) = (2(sinx + 1/2))/(2sinx)$

$2 (1 - {sin}^{2} x) = \frac{sin x + \frac{1}{2}}{sin x}$ $2(1 - sin^2x) = (sinx + 1/2)/(sinx)$

$(2 - 2 {sin}^{2} x) sin x = sin x + \frac{1}{2}$ $(2 - 2sin^2x )sinx = sinx + 1/2$

$2 sin x - 2 {sin}^{3} x = sin x + \frac{1}{2}$ $2sinx - 2sin^3x = sinx + 1/2$

$- 2 {sin}^{3} x + sin x - \frac{1}{2} = 0$ $-2sin^3x + sinx - 1/2 = 0$

We let $t = sin x$ $t= sinx$ .

$- 2 t^{3} + t - \frac{1}{2} = 0$ $-2t^3 + t - 1/2 = 0$

Solve using a graphing calculator, to get $t ≅ - 0.885$ $t ~= -0.885$ .

$sin x = - 0.885$ $sinx= -0.885$

$x = π + 2 π n + arcsin (0.885)$ $x = pi + 2pin+ arcsin(0.885)$ and $2 π n - arcsin (0.885)$ $2pin - arcsin(0.885)$

Hopefully this helps!

What is the general solution to the equation sin(3x) - sinx = 1sin(3x)−sinx=1sin(3x) - sinx = 1?