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JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

Find the number of solution of the equation 2^ ^2 x +5 · 2^ ^2 x =7...

Find the number of solution of the equation

2^{\sin ^2 x}+5 \cdot 2^{\cos ^2 x}=7

in the interval

[0,315]

.

Correct answer

100

Step-by-step solution

\begin{aligned} & 2^{\sin ^2 x}+5 \cdot 2^{\cos ^2 x}=7 \\ & 2^{\sin ^2 x}+5 \cdot 2^{1-\sin ^2 x}=7 \\ & 2^{\sin ^2 x}=t \Rightarrow t+\frac{10}{t}=7\end{aligned}

\begin{aligned} & \Rightarrow t^2-7 t+10=0 \Rightarrow(t-5)(t-2)=0 \Rightarrow t=2 \text { or } 5 \\ & \Rightarrow t=2^{\sin ^2 x}=2 \text { or } 5\end{aligned}

but

2^{\sin ^2 x} \neq 5

(think !)

\therefore \quad 2^{\sin ^2 x}=2=2^1 \Rightarrow \sin ^2 x=1

Hence values of x in

[0,315]

is

\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots . \frac{199 \pi}{2}

\therefore

total number of solution in

[0,315]

is 100

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