Question

How do I solve the trigonometric equation ${{\cos }^{2}}x={{\sin }^{2}}\left( \dfrac{x}{2} \right)$?

Answer 1

We need to find the correct formula to solve the given equation. After finding the formula, we should expand the formula. The expanded formula should be converted into a quadratic equation. Now solve the converted quadratic equation and find the roots. The resultant values are the solution for the ${{\cos }^{2}}x={{\sin }^{2}}\left( \dfrac{x}{2} \right)$.

Complete step-by-step solution:
The given equation is ${{\cos }^{2}}x ={{\sin }^{2}}\left( \dfrac{x}{2} \right)$
For us to solve the above equation. We can use the formula of sine half angle.
The formula is
$\Rightarrow \sin \left( \dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos \theta }{2}}$
As the formula is inside the square root, the plus or minus sign doesn’t make any changes for the result.
As we used the formula we should expand everything and later on, we must solve$\cos x$.
$\Rightarrow {{\cos }^{2}}x={{\sin }^{2}}\left( \dfrac{x}{2} \right)$
The square is applicable for both sin and value present in the bracket.
$\Rightarrow {{\cos }^{2}}x={{\left( \sin \left( \dfrac{x}{2} \right) \right)}^{2}}$
Now we need to substitute the $\left[ \sin \left( \dfrac{\theta }{2} \right)=\pm \sqrt{\dfrac{1-\cos \theta }{2}} \right]$ in the above equation
$\Rightarrow {{\cos }^{2}}x={{\left( \pm \sqrt{\dfrac{1-\cos x}{2}} \right)}^{2}}$
Here the square and the square root will get canceled.
$\Rightarrow {{\cos }^{2}}x=\dfrac{1-\cos x}{2}$.
$\Rightarrow 2{{\cos }^{2}}x=1-\cos x$.
Now we must change the above equation into a quadratic equation and we get
$\Rightarrow 2{{\cos }^{2}}x+\cos x-1=0$.
Now let’s solve this quadratic equation
$\Rightarrow 2{{\cos }^{2}}x+2\cos x-\cos x-1=0$.
Now we need to find the roots for the above quadratic equation.
$\Rightarrow 2\cos x\left( \cos x+1 \right)-1\left( \cos x+1 \right)=0$
$\Rightarrow \left( 2\cos x-1 \right)\left( \cos x+1 \right)=0$
Now we should take the term $2\cos x-1=0$
$\Rightarrow 2\cos x=1$
$\Rightarrow \cos x=\dfrac{1}{2}$
Now let’s take the other term $\cos x+1=0$
$\Rightarrow \cos x=-1$
Therefore the roots of the equation $2{{\cos }^{2}}x+2\cos x-\cos x-1=0$ are $\cos x=\dfrac{1}{2}$,$\cos x=-1$.
The cosine values for the $\dfrac{1}{2}$ are $\cos \dfrac{\pi }{3}$ and $\cos \dfrac{5\pi }{3}$
The cosine values $-1$ is $\cos {\pi}$
Hence the solution for ${{\cos }^{2}}x={{\sin }^{2}}\left( \dfrac{x}{2} \right)$ is $x=\dfrac{\pi }{3},\pi ,\dfrac{5\pi }{3}$.

Note: Double angle formula and half-angle formula are two types of angle formulas. We used the sine half-angle formula to solve the given equation in equation. This half-angle is used to evaluate the higher angles. For sin, cos, tan double angle formula is used. For sin, cos, tan, and squares of sin, the cos half-angle formula is used.