Question

4cos x.cos 2x.cos3x=1

Answer 1

Answer:

4cos x.cos 2x.cos3x=1

Step-by-step explanation:

4cosxcos2xcos3x=1

2cos2x(cos4x+cos2x)=1

2cos2x(2cos22x−1+cos2x)=1

4cos32x+2cos22x−2cos2x−1=0

2cos22x(2cos2x+1)−(2cos2x+1)=0

(2cos22x−1)(2cos2x+1)=0

cos22x=12 or cos2x=−12

cos2x=±12–√ or cos2x=−12

2x=nπ±π4 or (2n+1)π±π3

x=(4n±1)π8 or (2n+1)π2±π6,where n∈Z