Question

How do you solve $\cos 2x + {\sin ^2}x = 0$ from $0$ to $2\pi$?

Answer 1

Hint : This problem is related to trigonometry and the identities of it. Here we will use one of the various forms of $\cos 2x$ . then on simplifying we will get the equation in sin function form. That can be easily solved for the value of x. It is important to select the values within the range.

Complete step by step solution:
Given the function is,
$\cos 2x + {\sin ^2}x = 0$
Now the double angle of cos function can be written as $\cos 2x = 1 - 2{\sin ^2}x$
Thus the equation becomes,
$\Rightarrow 1 - 2{\sin ^2}x + {\sin ^2}x = 0$
$\Rightarrow 1 - {\sin ^2}x = 0$
Transposing sine function,
$\Rightarrow 1 = {\sin ^2}x$
Taking square root from both the sides,
$\Rightarrow \sin x = \pm 1$
Now ae we know x is from $0\;to\;2\pi$
Thus the value of the sin function for a right angle is +1. And sin function is negative in the third quadrant.
Thus $x = \dfrac{\pi }{2}\;or\;x = \dfrac{{3\pi }}{2}$
So, the correct answer is “$x = \dfrac{\pi }{2}\;or\;x = \dfrac{{3\pi }}{2}$”.

Note : Note that trigonometry related problems are very simple to solve. The only basic thing you should be prompt at is formulas, identities and value of angles or at least basic angles. Once we come to the point where every term or every equation is in sorted form then we can easily proceed to the problem.
Also note the ASTC rule that tells which function is positive and negative in which quadrant.
i.In the first quadrant All are positive.
ii.In the second quadrant only sin is positive.
iii.In the third quadrant only tan is positive.
iv.In the fourth quadrant only cos is positive.