Question

Find x.
cos(x-120) = sin2x
Final answer in rad.

Answer 1

Have you considered expansion by identity?

\(\cos(a-b) = \cos(a)\cos(b)+\sin(a)\sin(b)\)

\(\sin(2x) = 2\sin(x)\cos(x)\)

Answer 2

I got lost after \[- cosx - \sqrt{3}sinx = 4sinxcosx\]

Answer 3

We'll get back to the algebra in a minute. Let's check the trigonometry, first.

\(\cos(x-120º) = \cos(x)\cos(2\pi/3) + \sin(x)\sin(2pi/3) = (-1/2)\cos(x) + (\sqrt{3}/2)\sin(x)\)

Whoops! Looks like we missed a sign, in there.

This gets us to where you were:

\(-\cos(x) + \sqrt{3}\sin(x) = 4\sin(x)\cos(x)\)

Answer 4

Not really seeing how to go about that analytically.

Everything has period \(2\pi\). That simplifies things.

Various numerical methods can be employed to drag out the solutions.

I get, for k an integer:

\(x = 1.2217304764 + 2k\pi\)
\(x = 3.3161255788 + 2k\pi\)
\(x = 5.4105206812 + 2k\pi\)
\(x = 5.7595865316 + 2k\pi\)

Somewhat to my surprise, these could be:

\(x = 7\pi/18 + 2k\pi\)
\(x = 19\pi/18 + 2k\pi\)
\(x = 31\pi/18 + 2k\pi\)
\(x = 11\pi/6 + 2k\pi\)

So then, how can we solve analytically? Ideas?