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

Mabel Breault 2021-12-30 Answered
I want to solve for x
2sin4xcos2x2cos4xsin2x=cos2x
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Expert Answer

David Clayton
Answered 2021-12-31 Author has 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
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Donald Cheek
Answered 2022-01-01 Author has 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
[sin4xcos2x][cos4xsin2x]=(cos2xsin2x)(cos2x+sin2x+1)=2cos2x
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
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nick1337
Answered 2022-01-08 Author has 510 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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