Search any question & find its solution
Question:
Answered & Verified by Expert
If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$, then
Options:
Solution:
2117 Upvotes
Verified Answer
The correct answers are:
$\tan ^2 x=\frac{2}{3}$
,
$\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$\tan ^2 x=\frac{2}{3}$
,
$\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$$
\begin{aligned}
& \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \\
& \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \\
& \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\
& \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\
& \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \\
& \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125} \\
&
\end{aligned}
$$
\begin{aligned}
& \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \\
& \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \\
& \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\
& \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\
& \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \\
& \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125} \\
&
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.