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.