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Question : 106 of 150
Marks:
+1,
-0
Solution:
Given,
cos‌3‌x‌cos‌2‌x‌cos‌x= ⇒2(2‌cos‌x‌cos‌3‌x)‌cos‌2‌x=1 ⇒2(cos‌4‌x+cos‌2‌x)‌cos‌2‌x=1 [∵2‌cos‌A‌cos‌B=cos(A+B)+cos(A−B)] ⇒2(2cos22x−1+cos‌2‌x)‌cos‌2‌x=1 ⇒4cos32x−2‌cos‌2‌x +2cos22x−1=0 ⇒2‌cos‌2‌x(2cos22x−1) +1(2cos2x−1)=0 ⇒(2cos22x−1)(2‌cos‌2‌x+1)=0 ⇒cos‌4‌x(2‌cos‌2‌x+1)=0 ⇒cos‌4‌x=0 or
2‌cos‌2‌x+1=0 ⇒cos‌4‌x=cos‌ or
cos‌2‌x= ⇒4x= or
cos‌2‌x=cos‌ ⇒
x= or
2x= ⇒
x= or
x= Since,
x∈(0,) ∴x= is the required value.
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