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

Question

3174 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T14:36:10

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

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

$sin2xcosx + cos2xsinx - sinx = 1$

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

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

$4sinxcos^2x - 2sinx = 1$

$2sinx(2cos^2x- 1) = 1$

$2cos^2x - 1 = 1/(2sinx)$

$2cos^2x = 1+ 1/(2sinx)$

$2(1 - sin^2x) = (2sinx + 1)/(2sinx)$

$2 (1 - sin^2x) = (2(sinx + 1/2))/(2sinx)$

$2(1 - sin^2x) = (sinx + 1/2)/(sinx)$

$(2 - 2sin^2x )sinx = sinx + 1/2$

$2sinx - 2sin^3x = sinx + 1/2$

$-2sin^3x + sinx - 1/2 = 0$

We let $t= sinx$ .

$-2t^3 + t - 1/2 = 0$

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

$sinx= -0.885$

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

Hopefully this helps!

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