If f(x)= ^2 x+ ^2 2 x+ ^2 3 x , then the number of values of x ∈[0,2 π] for which f(x)=1 is

Correct answer: B. 6 aligned & f(x)= ^2 x+ ^2 2 x+ ^2 3 x=1 & ^2 x+(2 ^2 x-1)^2+(4 ^3 x-3 x)^2=1 & ^2 x+4 ^4 x+1-4 ^2 x+16 ^6 x & +9 ^2 x-24 ^4 x=1 & 6 ^2 x-20 ^4 x+16 ^6 x=0 & 2 ^2 x(3-10 ^2 x+8 ^4 x)=0 & ^2 x=0 or 8 ^4 x-10 ^2 x+3=0 & ^2 x=0 or (4 ^2 x-3)(2 ^2 x-1)=0 & ^2 x=0 or x=(3)/(4) or ^2 x=(1)/(2) & ^2 x=0 or x= 3 2 or = 1 2 & ^2 x=(2 n+1) (π)/(2), & x=2 n π ± (π)/(6) & aligned or array ll or & x=2 n π+(π)/(4) & x=(π)/(2), (3 π)/(2), (π)/(4), (π)/(6), (7 π)/(4), (11 π)/(6) array Hence, there are six values of x for which x ∈[0,2 π] .

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If f(x)= ^2 x+ ^2 2 x+ ^2 3 x , then the number of values of x...

f(x)=\cos ^2 x+\cos ^2 2 x+\cos ^2 3 x

, then the number of values of

x \in[0,2 \pi]

for which

f(x)=1

Options

Correct answer

B. 6

Step-by-step solution

\begin{aligned}& f(x)=\cos ^2 x+\cos ^2 2 x+\cos ^2 3 x=1 \\ & \Rightarrow \cos ^2 x+\left(2 \cos ^2 x-1\right)^2+\left(4 \cos ^3 x-3 \cos x\right)^2=1 \\ & \Rightarrow \cos ^2 x+4 \cos ^4 x+1-4 \cos ^2 x+16 \cos ^6 x \\ & +9 \cos ^2 x-24 \cos ^4 x=1 \\ & \Rightarrow \quad 6 \cos ^2 x-20 \cos ^4 x+16 \cos ^6 x=0 \\ & \Rightarrow \quad 2 \cos ^2 x\left(3-10 \cos ^2 x+8 \cos ^4 x\right)=0 \\ & \Rightarrow \quad \cos ^2 x=0 \text { or } 8 \cos ^4 x-10 \cos ^2 x+3=0 \\ & \Rightarrow \quad \cos ^2 x=0 \text { or }\left(4 \cos ^2 x-3\right)\left(2 \cos ^2 x-1\right)=0 \\ & \Rightarrow \quad \cos ^2 x=0 \text { or } \cos x=\frac{3}{4} \text { or } \cos ^2 x=\frac{1}{2}\\ & \Rightarrow \quad \cos ^2 x=0 \text { or } \cos x=\frac{\sqrt{3}}{2} \text { or } \cosx=\frac{1}{\sqrt{2}}\\ & \Rightarrow \quad \cos ^2 x=(2 n+1) \frac{\pi}{2},\\ & x=2 n \pi \pm \frac{\pi}{6}\\ & \end{aligned}

\begin{array}{ll} \text { or }& x=2 n \pi+\frac{\pi}{4}\\ \therefore & x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}, \frac{7 \pi}{4}, \frac{11 \pi}{6} \end{array}

Hence, there are six values of

x

for which

x \in[0,2 \pi]

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