Question

Find the general solution of $|\sin x| = \cos x$ is (when $n \in I$) given by
A) $n\pi + \dfrac{\pi }{4}$
B) $2n\pi \pm \dfrac{\pi }{4}$
C) $n\pi \pm \dfrac{\pi }{4}$
D) $n\pi - \dfrac{\pi }{4}$

Answer 1

The trigonometric functions are functions which relate the angle of the right angled triangle to the ratios of two side lengths . There are six trigonometric functions , there are sine , cosine , tangent , cosecant , secant and cotangent .
Formula between sine and cosine is ${\sin ^2}x + {\cos ^2}x = 1$ . In any case a variable positive or negative we get the mod value i.e., $| + a| = | - a| = a$ , where $a$ is a variable, Square of any modulus begins positive terms .

Complete step by step answer:
Given $|\sin x| = \cos x$ , $x \in I$
Squaring both sides we get ,
${(|\sin x|)^2} = {(\cos x)^2}$
$\Rightarrow {\sin ^2}x = {\cos ^2}x$
From the formula of sine and cosine ${\sin ^2}x + {\cos ^2}x = 1$
Using Pythagorean identity of right angled triangle we will have ${\sin ^2}x = 1 - {\cos ^2}x$
We substitute this in above equation we will have
$\Rightarrow 1 - {\cos ^2}x = {\cos ^2}x$
$\Rightarrow 2{\cos ^2}x = 1$
Dividing both sides by $2$ , we get
$\Rightarrow {\cos ^2}x = \dfrac{1}{2}$
We know $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
Therefore $\dfrac{1}{2} = {\cos ^2}\dfrac{\pi }{4}$ , put this and we get
$\Rightarrow {\cos ^2}x = {\cos ^2}\dfrac{\pi }{4}$
Taking square root of both sides , we get
$\Rightarrow \cos x = \cos \dfrac{\pi }{4}$ ………………………………..(i)
If $\cos x = \cos y$ then we get the general solution $x = 2n\pi \pm y$
$\Rightarrow x = 2n\pi \pm \dfrac{\pi }{4}$ , in this case $y = \dfrac{\pi }{4}$
Therefore, option (B) is correct.

Note:
In the equation (i) , we get always positive value because given in the question $\cos x = |\sin x|$ , for this reason we do not take the negative value of x and in other way we know $\cos x = \cos ( - x)$ . So the option (B) is correct .
We can also solve the given problem by using the like this we will put this formula ${\cos ^2}x - {\sin ^2}x = \cos 2x$ in the given equation and you get the required answer .