Question

Solve the equation $\cos 2x = 0$.

Answer 1

Hint: In this question, we need to find a general solution of cos2x = 0. For this, we will use the general form of cosx according to which if $\cos \theta =\cos \alpha $ then $\theta =2n\pi \pm \alpha $. Here we will first find the value of $\alpha $ such that $\cos \alpha =0$ and equate it to cos2x, to find the value of 2x in terms of found $\alpha $. Then we will divide by 2 to get a final solution.

Complete step-by-step solution
Here we are given an equation as cos2x = 0. We need to find the value of x. Since no interval is given, so we need to find the value of x in general (in terms of n where n = 0, 1, 2. . . . . . . .)
As we know that, the general form of cosx states that if $\cos \theta =\cos \alpha $ then, $\theta =2n\pi \pm \alpha $ where n = 0, 1, 2 . . . .
So let us find values of $\alpha $ such that $\cos \alpha =0$.
From known values of cosine function, we know that $\cos \dfrac{\pi }{2}=0$. Hence, $\alpha =\dfrac{\pi }{2}$. So we get:
$\cos 2x=\cos \dfrac{\pi }{2}$ which implies that $2x=2n\pi \pm \dfrac{\pi }{2}$ where n = 0, 1, 2 . . . . .
We have found value of 2x but we need the value of x, so dividing both sides by 2, we get:
$\dfrac{2x}{2}=\dfrac{2n\pi \pm \dfrac{\pi }{2}}{2}$ where n = 0, 1, 2. . . . . .
Cancelling 2 on the left side of the equation and separating terms on the right side of the equation, we get:
$x=\dfrac{2n\pi }{2}\pm \dfrac{\pi }{4}$ where n = 0, 1, 2. . . . . . . . .
$\Rightarrow x=n\pi \pm \dfrac{\pi }{4}$ where n = 0, 1, 2. . . . . . . . .
Now taking $\pi $ common from both terms on the right side of the equation, we get:
$x=\left( n\pm \dfrac{1}{4} \right)\pi $ where n = 0, 1, 2. . . . . . . .
Hence this is our required solution of cos2x = 0.

Note: Students should note that, we have to start taking values from n = 0. They can check their answer by putting the value of n and checking if cos2x = 0. For example, for n = 1, we have $x=\left( 1\pm \dfrac{1}{4} \right)\pi \Rightarrow x=\dfrac{3}{4}\pi ,\dfrac{5}{4}\pi $. Cos2x becomes $\cos 2\left( \dfrac{3}{4}\pi \right)\text{ and }\cos 2\left( \dfrac{5}{4}\pi \right)$.
Now, $\cos \dfrac{3}{2}\pi =\cos \left( \pi +\dfrac{\pi }{2} \right)=\cos \dfrac{\pi }{2}=0$.
Also, $\cos \dfrac{5}{2}\pi =\cos \left( 2\pi +\dfrac{\pi }{2} \right)=\cos \dfrac{\pi }{2}=0$
Don't forget to take both positive and negative signs in the general form of the value of the cosine function. Students should know general solutions of all the trigonometric functions to solve these sums.