Question

If $\frac{si{n}^{4}x}{2}+\frac{co{s}^{4}x}{3}=\frac{1}{5}$ then

• A. $ta{n}^{2}x=\frac{2}{3}$
• B. $\frac{si{n}^{8}x}{8}+\frac{co{s}^{8}x}{27}=\frac{1}{125}$
• C. $ta{n}^{2}x=\frac{1}{3}$
• D. $\frac{si{n}^{8}x}{8}+\frac{co{s}^{8}x}{27}=\frac{2}{125}$

Answer 1

Answer: (A), (B)

$\frac{si{n}^{4}x}{2}+\frac{co{s}^4}{3}=\frac{1}{5}$
$\Rightarrow \frac{si{n}^{4}x}{2}+\frac{(1-si{n}^{2}x{)}^2}{3}=\frac{1}{5}$
$\Rightarrow \frac{si{n}^4}{2}+\frac{1+si{n}^{4}x-2si{n}^{2}x}{3}=\frac{1}{5}$
$\Rightarrow 5si{n}^{4}x-4si{n}^{2}x+2=\frac{6}{5}$
$\Rightarrow 25si{n}^{4}x-20si{n}^{2}x+4=0$
$\Rightarrow (5si{n}^{2}x-2{)}^2$=0
$\Rightarrow si{n}^{2}x=\frac{2}{5}$
$si{n}^{2}x=\frac{3}{5},ta{n}^{2}x=\frac{2}{3}$
$\frac{si{n}^{8}x}{8}+\frac{co{s}^{8}x}{27}=\frac{1}{125}$