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Question

Question: Find the general solution of the equation \({\sec ^2}2x = 1 - \tan 2x\)...

Find the general solution of the equation
sec22x=1tan2x{\sec ^2}2x = 1 - \tan 2x

Explanation

Solution

Hint - Express sec2θ=1+tan2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta and solve the problem.

Complete step-by-step answer:

The given equation is sec22x{\sec ^2}2x
So, if we express in terms of tan , it would be equal to
1+tan22x1 + {\tan ^2}2x =1-tan2x
On shifting and rearranging the terms, we get this to be equal to
tan2x(tan2x+1)=0
So, from this we get the value of tan2x=0 or tan2x=-1
Shifting tan to the other side, we get
2x=tan10\Rightarrow 2x = {\tan ^{ - 1}}0 or 2x=tan1(1)2x = {\tan ^{ - 1}}( - 1)
2x=nπ,nππ4\Rightarrow 2x = n\pi ,n\pi - \dfrac{\pi }{4}
So, from this we get the general solution of the equation, that is
x=nπ2,nπ2π8x = \dfrac{{n\pi }}{2},\dfrac{{n\pi }}{2} - \dfrac{\pi }{8}
So, this is the general solution of the equation.

Note: In accordance to the value which has to be found , make use of the appropriate trigonometric identities and solve this type of problem and also express the equation given in a convenient form before solving it further.