Step 1: Simplification
Given: cos4x=cos2x
⇒cos4x−cos2x=0
⇒−2sin(4x+2x2)sin(4x−2x2)=0 [∵cosA−cosB=−2sin(A+B2)sin(A−B2)]
⇒−2sin(6x2)sin(2x2)=0
⇒−2sin3xsinx=0
⇒sin3xsinx=0
So, either sin3x=0 or sinx=0
We solve sin3x=0 and sinx=0 separately.
Step 2: General solution for sin3x=0
sin3x=0 or sin3x=sin0
So the general solution is
3x=nπ±(−1)n(0)
⇒3x=nπ
⇒x=nπ3
Step 3: General solution for sinx=0
sinx=0
⇒x=nπ
Final answer :x=nπ, nπ3