I want to solve for x 2sin4x−cos2x−2cos4x−sin2x=cos⁡2x

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I want to solve for x  2sin4x−cos2x−2cos4x−sin2x=cos⁡2x

David Clayton · Accepted Answer

Let u=sin2(x) and v=cos2(x) , then 2u2−v−2v2−u=v−u and u+v=1. Thus 2u2+u−1+u=2v2+v−1+vDefine f(u)=2u2+u−1 ,then we are looking for a u∈[0,1] such that f(u)=f(1-u). However f′(u)=ln⁡(2)(2u+1)2u2+u−1+1&amp;gt;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 · Accepted Answer

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

[\sin^{4} x - \cos^{2} x] - [\cos^{4} x - \sin^{2} x] = - (\cos^{2} x - \sin^{2} x) (\cos^{2} x + \sin^{2} x + 1) = - 2 \cos 2 x

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 = \pm \frac{π}{4} + 2 π Z, \pm \frac{3 π}{4} + 2 π Z

nick1337 · Accepted Answer

Putting cos2⁡x=a sin4⁡x−cos2⁡x=(1−a)2−a=a2−3a+1 and cos4⁡x−sin2⁡x=a2−(1−a)=a2+a−1 So we get, 2a2−3a+1−2a2+a−1=2a−1 or, 2a2−3a+1(1−24a−2)=2a−1 If a−1&amp;gt;0,  4a−2&amp;gt;0⇒24a−2&amp;gt;20=1⇒ the left hand side is &amp;lt; 0 Similarly, if 2a-1 &amp;lt;0 the right hand side is &amp;gt; 0 If 2a-1=0, both sides become 0 ⇒2cos2⁡x−1=0⇔cos⁡2x=0⇒2x=(2n+1)π2 where n is any integer

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

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