\(5\sin 2x^\circ + 7\cos x^\circ = 0\)

Replace \(\sin 2x\) with \(2\sin x\cos x\)

\(5(2\sin x^\circ \cos x^\circ ) + 7\cos x^\circ = 0\)

\(10\sin x^\circ \cos x^\circ + 7\cos x = 0\)

Take out \(\cos x^\circ\) as the common factor, but don't divide by the common factor.

\(\cos x^\circ (10\sin x^\circ + 7) = 0\)

So either \(\cos x^\circ = 0\) or \(10\sin x^\circ + 7 = 0\)

\(\cos x^\circ = 0\)

\(x^\circ = 90^\circ \,or\,270^\circ\)

\(10\sin x^\circ + 7 = 0\)

\(\sin x^\circ = - \frac{7}{{10}}\)

\(x^\circ = 224.4^\circ \,or\,315.6^\circ\)

Which gives us solutions of \(90^\circ ,224.4^\circ ,270^\circ ,315.6^\circ\)

There are two more methods for you to learn, both of which involve \(\cos 2x\).

• For any trigonometric equation involving \(\cos 2x\) and \(\sin x\), replace \(\cos 2x\) with \(1 - 2{\sin ^2}x\)
• For any trigonometric equation involving \(\cos 2x\) and \(\cos x\), replace \(\cos 2x\) with \(2{\cos ^2}x - 1\)

Then, in either case, take all the terms to one side, creating a quadratic equation in terms of \(\sin x\) or \(\cos x\) only and then solve.

\(\cos 2x + 3\sin x + 1 = 0\)

\(1 - 2{\sin ^2}x + 3\sin x + 1 = 0\)

\(- 2{\sin ^2}x + 3\sin x + 2 = 0\)

\(2{\sin ^2}x - 3\sin x - 2 = 0\)

\((2\sin x + 1)(\sin x - 2) = 0\)

\(2\sin x + 1 = 0\)

\(\sin x - 2 = 0\)

\(\sin x = - \frac{1}{2}\)

\(\sin x = 2\)

Since \(- 1 \le \sin x \le 1\), this equation has no solutions, which gives:

\(x = \frac{{7\pi }}{6}\) or \(\frac{{11\pi }}{6}\)