How to Solve Trigonometric Equations Using Identities

$$\begin{array}{llll}\hfill 4\cos 2x+4\sin 2x& =0|÷4& \hfill & \\ \hfill \cos 2x+\sin 2x& =0|÷\cos 2x\ne 0& \hfill & \\ \hfill \frac{\cos 2x}{\cos 2x}+\frac{\sin 2x}{\cos 2x}& =0& \hfill & \\ \hfill 1+\tan 2x& =0& \hfill & \\ \hfill \tan 2x& =-1& \hfill & \\ \hfill 2x& =\frac{3\pi }{4}+n\cdot \pi & \hfill & \\ \hfill x& =\frac{3\pi }{8}+n\cdot \frac{\pi }{2}& \hfill & \end{array}$$ In the interval $\left[0,4\pi \right)$, that gives you the solutions Now you have to check if $\cos 2x=0$ gives any solutions. Note that $\cos 2x=0$ is the same as $\sin 2x=1$ or $\sin 2x=-1$: From this you get the following solutions in the interval $\left[0,4\pi \right)$:

$$x\in \left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4},\frac{13\pi }{4},\frac{15\pi }{4}\right\}$$

Now you have to test these answers in the main equation (1):

$x={\displaystyle \frac{\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{\pi }{4}\right)+4\sin \left(2\cdot \frac{\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{3\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{3\pi }{4})\right)+4\sin \left(2\cdot \frac{3\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{5\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{5\pi }{4}\right)+4\sin \left(2\cdot \frac{5\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{7\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{7\pi }{4}\right)+4\sin \left(2\cdot \frac{7\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{9\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{9\pi }{4}\right)+4\sin \left(2\cdot \frac{9\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{11\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{11\pi }{4}\right)+4\sin \left(2\cdot \frac{11\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{13\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{13\pi }{4}\right)+4\sin \left(2\cdot \frac{13\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

$x={\displaystyle \frac{15\pi }{4}}$:

$$\begin{array}{llll}\hfill \text{LHS}& =4\cos \left(2\cdot \frac{15\pi }{4}\right)+4\sin \left(2\cdot \frac{15\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

Equation (1) doesn’t have any more solutions, and you can conclude that the solutions in the interval $\left[0,4\pi \right)$ are:

Note! This is a very cumbersome way to check whether $\cos 2x=0$ yields multiple solutions, but it always works! Another way of checking $\cos 2x=0$ is to look at the main equation and using what you know about trigonometric identities to analyse it.

$$\begin{array}{llll}\hfill \cos (2x)+\cos x+1& =0& \hfill & \\ \hfill {\cos }^{2}x-{\sin }^{2}x+\cos x+1& =0& \hfill & \\ \hfill {\cos }^{2}x-\left(1-{\cos }^{2}x\right)+\cos x+1& =0& \hfill & \\ \hfill 2{\cos }^{2}x+\cos x& =0& \hfill & \\ \hfill \cos x(2\cos x+1)& =0& \hfill & \end{array}$$ You use the zero product property and look at the two equations $\cos x=0$ and $2\cos x+1=0$. The equation $\cos x=0$ has the solutions $$\begin{array}{llll}\hfill x_1& =\frac{\pi }{2}+n\cdot 2\pi & \hfill & \\ \hfill x_2& =-\frac{\pi }{2}+n\cdot 2\pi & \hfill & \end{array}$$

You rewrite the equation $2\cos x+1=0$ as the basic equation $\cos x=-\frac{1}{2}$. It has the solutions

$$\begin{array}{llll}\hfill x_3& =\frac{2\pi }{3}+n\cdot 2\pi & \hfill & \\ \hfill x_4& =-\frac{2\pi }{3}+n\cdot 2\pi & \hfill & \end{array}$$

That means the solutions for the interval $\left[0,2\pi \right)$ are

$$x\in \left\{\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2}\right\}$$

$$\begin{array}{llll}\hfill 2\left(1-{\sin }^{2}x\right)-\sin x& =1& \hfill & \\ \hfill 2-2{\sin }^{2}x-\sin x& =1& \hfill & \\ \hfill -2{\sin }^{2}x-\sin x+1& =0& \hfill & \end{array}$$ You make the substitution $u=\sin x$ in this equation, which gives you

$$-2u^2-u+1=0.$$

Solve the quadratic equation:

The solutions to this are

$$u_1=-1u_2=\frac{1}{2}$$

Now you substitute $\sin x$ back in for $u$:

The solutions in the interval $\left[0,2\pi \right)$ are:

$$x\in \left\{\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}\right\}$$

Note that $-\frac{\pi }{2}$ is not included among the solutions. That’s because it’s outside the interval we’re looking at.