Question

How do you solve $\cos \left(2x\right)=-\frac{\sqrt{3}}{2}$ using the double angle identity?

Answer 1

Step- 1: Express $-\frac{\sqrt{3}}{2}$ in terms of cosine of some angle:

$\begin{array}{c}\begin{array}{rcll}-\frac{\sqrt{3}}{2}& =& -\cos \left(30°\right)& \left(\text{substituted using known value}\cos \left(30°\right)\mathrm{=}\frac{\sqrt{3}}{2}\right)\\ & =& \cos \left(180°-30°\right)& \left(\text{Applied the identity}\cos \left(180°-x\right)=-\cos \left(x\right)\right)\\ & =& \cos \left(150°\right)& \end{array}\end{array}$

Thus, $-\frac{\sqrt{3}}{2}=\cos \left(150°\right)$. So, the given equation can be written as $\cos \left(2x\right)=\cos \left(150°\right)$.

Step- 2: Simplify the equation after using the double angle identity for cosine:

$\begin{array}{lrclll}& \cos \left(2x\right)& =& \cos \left(150°\right)& & \\ \Rightarrow & 2{\left(\cos \left(x\right)\right)}^2-1& =& 2{\left(\cos \left(75°\right)\right)}^2-1& & \left(\text{Applied the double angle identity}\cos \left(2x\right)=2{\left(\cos \left(x\right)\right)}^2-1\right)\\ \Rightarrow & 2{\left(\cos \left(x\right)\right)}^2& =& 2{\left(\cos \left(75°\right)\right)}^2& & \\ \Rightarrow & {\left(\cos \left(x\right)\right)}^2& =& {\left(\cos \left(75°\right)\right)}^2& & \\ \Rightarrow & \begin{array}{rr}{\left(\cos \left(x\right)\right)}^2-& {\left(\cos \left(75°\right)\right)}^2\end{array}& =& 0& & \\ \Rightarrow & \left(\cos \left(x\right)-\cos \left(75°\right)\right)\left(\cos \left(x\right)+\cos \left(75°\right)\right)& =& 0& & \left(\text{Applied the identity}{a}^2-b^2=\left(a+b\right)\left(a-b\right)\right)\\ \Rightarrow & \cos \left(x\right)& =& \pm \cos \left(75°\right)& & \end{array}$

Thus, $\begin{array}{rcl}\cos \left(x\right)& =& \pm \cos \left(75°\right)\end{array}$.

Step- 3: Solve for $x$ using $\begin{array}{rcl}\cos \left(x\right)& =& \pm \cos \left(75°\right)\end{array}$:

Recall that if $\cos \left(x\right)=\cos \left(y\right)$ then $x=n\times 360°\pm y$, where $n$ is any integer.

Since$\begin{array}{rcl}\cos \left(x\right)& =& \cos \left(75°\right)\end{array}$, we get $x=n\times 360°\pm 75°$.

Also, $\begin{array}{l}\begin{array}{rcl}\cos \left(x\right)& =& -\cos \left(75°\right)\end{array}\end{array}$, thus

$\begin{array}{rcl}& & \begin{array}{rclll}\cos \left(x\right)& =& -\cos \left(75°\right)& & \\ & =& \cos \left(180°-75°\right)& & \left(\text{Applied}\cos \left(180°-x\right)=-\cos \left(x\right)\right)\\ & =& \cos \left(105°\right)& & \end{array}\end{array}$.

Since $\begin{array}{rcl}\cos \left(x\right)& =& \cos \left(105°\right)\end{array}$, we get $x=n\times 360°\pm 105°$.

Hence, $\mathit{x}\mathbf{=}\mathit{n}\mathbf{\times }\mathbf{360}\mathbf{°}\mathbf{\pm }\mathbf{75}\mathbf{°}$ and $\mathit{x}\mathbf{=}\mathit{n}\mathbf{\times }\mathbf{360}\mathbf{°}\mathbf{\pm }\mathbf{105}\mathbf{°}$ together, form the general solution of the given equation $\mathit{c}\mathit{o}\mathit{s}\left(2x\right)\mathbf{=}\mathbf{-}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$. Here $n$ is an integer.