Question

The solution of trigonometric equation ${\cos ^4}x + {\sin ^4}x = 2\cos (2x + \pi )\cos (2x - \pi )$ is
A.$x = \dfrac{{n\pi }}{2} \pm {\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right)$
B.$x = \dfrac{{n\pi }}{2} + \dfrac{{{{( - 1)}^n}}}{4}{\sin ^{ - 1}}\left( {\dfrac{{ \pm 2\sqrt 2 }}{3}} \right)$
C.$x = \dfrac{{n\pi }}{2} \pm {\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)$
D.$x = \dfrac{{n\pi }}{2} - \dfrac{{{{( - 1)}^n}}}{4}{\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)$

Answer 1

Hint : A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation . Solving an equation means to find the set of all the values of the unknown angle which satisfies the given equation . Since , the trigonometric functions are periodic therefore , a trigonometric equation has a solution it will have infinitely many solutions .

Complete step-by-step answer :
Given : ${\cos ^4}x + {\sin ^4}x = 2\cos (2x + \pi )\cos (2x - \pi )$
Using the identities of $2\cos C\cos D = \cos (A + B) + \cos (A - B)$ , in the LHS of the equation we get ,
${\cos ^4}x + {\sin ^4}x = \cos (2x + \pi + 2x - \pi ) + \cos (2x + \pi - 2x + \pi )$
On simplifying we get ,
${({\cos ^2}x)^2} + {({\sin ^2}x)^2} = \cos (4x) + \cos (2\pi )$
Now making adjustment on RHS we get
${({\cos ^2}x)^2} + {({\sin ^2}x)^2} + 2{\sin ^2}x{\cos ^2}x - 2{\sin ^2}x{\cos ^2}x = 2\cos (4x) + \cos (2\pi )$
on solving we get ,
${({\cos ^2}x + {\sin ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x = \cos (4x) + \cos (2\pi )$ ,
using the identity ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2\pi = 1$ in the above equation we get,
$1 - 2{\sin ^2}x{\cos ^2}x = \cos (4x) + 1$
Now using the formula $\sin 2x = 2\sin x\cos x$ , we get
$1 - \dfrac{1}{2}{\sin ^2}2x = \cos (4x) + 1$
On solving further we get ,
$- \dfrac{1}{2}{\sin ^2}2x = \cos (4x)$
Now using the formula $\cos 2A = 1 - 2{\sin ^2}A$ on RHS we get
$- \dfrac{1}{2}\left( {\dfrac{{1 - \cos 4x}}{2}} \right) = \cos (4x)$
Here we have got $\cos 4x$as we have ${\sin ^2}2x$ in the above equation
$1 - \cos 4x = - 4\cos (4x)$
$3\cos 4x = - 1$ , on solving further we get
$\cos 4x = \dfrac{{ - 1}}{3}$
$2{\cos ^2}2x - 1 = \dfrac{{ - 1}}{3}$ , on solving further we get
${\cos ^2}2x = \dfrac{1}{3}$
now using the identity of ${\sin ^2}x + {\cos ^2}x = 1$ , the value of ${\cos ^2}2x$ can be represented in terms of ${\sin ^2}2x$ . Therefore ,
${\sin ^2}2x = 1 - {\cos ^2}2x$
${\sin ^2}2x = 1 - \dfrac{1}{3}$
On solving further we get ,
${\sin ^2}2x = \dfrac{{3 - 1}}{3}$ ,
${\sin ^2}2x = \dfrac{2}{3}$
taking square root on both sides we get
$\sin 2x = \sqrt {\dfrac{2}{3}}$
Now using the general equation for formula for $\sin \theta = \sin \alpha$ , we have $\theta = n\pi + {( - 1)^n}\alpha ,n \in Z$
Therefore ,
$x = \dfrac{{n\pi }}{2} + {( - 1)^n}{\sin ^{ - 1}}\left( { \pm \sqrt {\dfrac{2}{3}} } \right),n \in Z$
Therefore ,none of the options is the correct answer for the given trigonometric equation .

Note : In the given question it has asked about the trigonometric solution, so the solution of any trigonometric equation is always given in the form of a general solution , which is different for different trigonometric equations . A solution generalized by means of periodicity is known as the general solution.