Question

If $4{\sin ^4}x + {\cos ^4}x = 1$, then $x$ is equal to ($n \in Z$)
This question has multiple correct answers.
${{(A) n\pi }}$
${{(B}}{{) n\pi }} \pm {\text{si}}{{\text{n}}^{ - 1}}(\sqrt {\dfrac{2}{5}} )$
${{(C) }}\dfrac{{2n\pi }}{3}$
${{(D) 2n\pi }} \pm \dfrac{\pi }{4}$

Answer 1

Hint: In this question we will try to simplify the equation given to us in the form of $ab = 0$ and using the formula. Finally we get the required answer of $x$.

Formula used: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
${\sin ^2}x + {\cos ^2}x = 1$
$1 - {\cos ^2}x = {\sin ^2}x$

Complete step-by-step solution:
From the question, it is given that the equation is:
$4{\sin ^4}x + {\cos ^4}x = 1$
On transferring ${\cos ^4}x$ across the $ = $ sign we get:
$\Rightarrow$ $4{\sin ^4}x = 1 - {\cos ^4}x$
Since the right-hand side in the above equation is in the form ${a^2} - {b^2}$ we can expand it as follows:
$\Rightarrow$$4{\sin ^4}x = (1 - {\cos ^2}x)(1 + {\cos ^2}x)$
Now we know that ${\sin ^2}x + {\cos ^2}x = 1$, therefore$1 - {\cos ^2}x = {\sin ^2}x$, substituting in the above equation we get:
$\Rightarrow$$4{\sin ^4}x = ({\sin ^2}x)(1 + {\cos ^2}x)$
On taking the right-hand side to the left-hand side we get:
$\Rightarrow$$4{\sin ^4}x - {\sin ^2}x(1 + {\cos ^2}x) = 0$
Now since both the terms have ${\sin ^2}x$ common, we remove it as common:
$\Rightarrow$${\sin ^2}x(4{\sin ^2}x - (1 + {\cos ^2}x)) = 0$
Now since the above equation is in the format $ab = 0$, either $a = 0$or$b = 0$,
So we can write it as,
$\Rightarrow$${\sin ^2}x = 0$ Or $4{\sin ^2}x - (1 + {\cos ^2}x) = 0$
Now if ${\sin ^2}x = 0$,
$\Rightarrow$$\sin x = 0$
And $\sin x = 0$ when $x = n\pi $
Therefore, one correct option is option $(A)$.
Again we can write the $4{\sin ^2}x - (1 + {\cos ^2}x) = 0$
We know that ${\cos ^2}x = 1 - {\sin ^2}x$ therefore the equation can be written as:
$4{\sin ^2}x - (1 + 1 - {\sin ^2}x) = 0$
On opening the bracket, we get:
$\Rightarrow$$4{\sin ^2}x - 2 + {\sin ^2}x = 0$
On transferring $2$ across the $ = $ sign we get:
$\Rightarrow$$4{\sin ^2}x + {\sin ^2}x = 2$
On simplifying we get:
$\Rightarrow$$5{\sin ^2}x = 2$
Therefore,
${\sin ^2}x = \dfrac{2}{5}$
On taking square root both the sides, we get:
$\Rightarrow$$\sin x = \sqrt {\dfrac{2}{5}} $
Therefore $\sin x = \sin \alpha $
$\Rightarrow$$x = n\pi + {( - 1)^n}{\sin ^{ - 1}}(\sqrt {\dfrac{2}{5}} )$
Therefore, the second correct option is $(B)$.

Therefore, options $(A)$ and option $(B)$ both are correct in this question.

Note: The formula used over here is for $\sin (n\pi + x)$,
It is to be remembered that $\sin (n\pi + x) = {( - 1)^n}\sin x$
Basic trigonometric formulas should be remembered to solve these types of sums.
The inverse trigonometric function of $\sin x$ which is ${\sin ^{ - 1}}x$ used in this sum
For example, if $\sin x = a$ and then $x = {\sin ^{ - 1}}a$.
Also, ${\sin ^{ - 1}}(\sin x) = x$ this is a property of the inverse function.
There also exists inverse functions for the other trigonometric relations such as $\tan $ and $\cos $.
The inverse function is used to find the angle $x$ from the value of the trigonometric relation.