If cos⁡x⋅cos⁡2x=14, x∈[0,90∘), then what is the solution of the equation? I attempted to solve this question as follows: cos⁡2x=cos2x−sin2x ⇒cos⁡x(cos2x−sin2x)=14 cos3x−sin⁡x(cos⁡xsin⁡x) And I got stuck here, I did not know how to continue.

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If cos⁡x⋅cos⁡2x=14, x∈[0,90∘), then what is the solution of the equation?   I attempted to solve this question as follows:  cos⁡2x=cos2x−sin2x  ⇒cos⁡x(cos2x−sin2x)=14  cos3x−sin⁡x(cos⁡xsin⁡x)  And I got stuck here, I did not know how to continue.

xandir307dc · Accepted Answer

Since x≠0, multiplying both sides by 4sin⁡x and using double angle formula, sin⁡2θ=2sin⁡θcos⁡θ twice, it is obtained  sin⁡4x=sin⁡x  whence  4x+x=π⇒x=π5=36∘

Annie Gonzalez · Accepted Answer

cos⁡x⋅cos⁡(2x)=14⇒4cos⁡x⋅(2cos2x−1)=1⇒8cos3x−4cos⁡x−1=0  Observe that cos⁡x=−12 is a solution. Can you finish it with factoring ?

Vasquez · Accepted Answer

cos⁡x(2cos2⁡x−1)=14=(−12)2→cos⁡x=−12=cos⁡(π−π3)→x=(2k+1)π−π3

If \cos x \cdot \cos 2x=\frac14, \ x \in [0,90^{\circ}),

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