I want to solve for x 2^{\sin^4 x-\cos^2 x}-2^{\cos^4 x-\sin^2 x}=\cos

Mabel Breault

Mabel Breault

Answered question


I want to solve for x

Answer & Explanation

David Clayton

David Clayton

Beginner2021-12-31Added 36 answers

Let u=sin2(x) and v=cos2(x) , then 2u2v2v2u=vu and u+v=1. Thus 2u2+u1+u=2v2+v1+v
Define f(u)=2u2+u1 ,then we are looking for a u[0,1] such that f(u)=f(1-u). However f(u)=ln(2)(2u+1)2u2+u1+1>0 for all u[0,1]. Hence f is injective on [0,1] and thus f(u)=f(1-u) if and only if u=1-u. Thus sin2(x)=u=12
Donald Cheek

Donald Cheek

Beginner2022-01-01Added 41 answers

The left side of the equation is greater than 0 if and only if the right side is less than 0, and vice versa.
This follows from
Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives x=±π4+2πZ,±3π4+2πZ


Expert2022-01-08Added 777 answers

Putting cos2x=a
sin4xcos2x=(1a)2a=a23a+1 and cos4xsin2x=a2(1a)=a2+a1
So we get, 2a23a+12a2+a1=2a1
or, 2a23a+1(124a2)=2a1
If a1>0,  4a2>024a2>20=1 the left hand side is < 0
Similarly, if 2a-1 <0 the right hand side is > 0
If 2a-1=0, both sides become 0
2cos2x1=0cos2x=02x=(2n+1)π2 where n is any integer

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