The general solution to sin 10 x -cos 10 x -29-16cos 4 2 x isA- x-n -4-8- n - IB- x-2 n -2- n - IC- x-n -2- n - ID- x-2 n -2- n - 1

The correct option is A sin^10 x + cos^10 x =29 16 cos^4 2x or, (1 cos2x 2)^5+(1+cos2x 2)^5=29 16cos^4 2x Let cos 2x=t then (1 t 2)^5+(1+t 2)^5=29 16t^4 or, 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2

The general s...

Question

The general solution to $s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$ is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A

$s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$
or, $(\frac{1 - c o s 2 x}{2})^{5} + (\frac{1 + c o s 2 x}{2})^{5} = \frac{29}{16} c o s^{4} 2 x$
Let cos 2x=t
then $(\frac{1 - t}{2})^{5} + (\frac{1 + t}{2})^{5} = \frac{29}{16} t^{4}$
or, $24 t^{4} - 10 t^{2} - 1 = 0$
or, $(2 t^{2} - 1) (12 t^{2} + 1) = 0$
or, $12 t^{2} + 1 \neq 0$
$∴ s o, 2 t^{2} - 1 = 0 t h e n t^{2} = \frac{1}{2}$
Put the value of t
so, $c o s^{2} 2 x = \frac{1}{2}$ or, $2 c o s^{2} 2 x - 1 = 0$
or, $c o s 4 x = 0$
or, $4 x = n π + \frac{π}{2}$ or, $x = \frac{n π}{4} + \frac{π}{8}, n \in I$

Suggest Corrections

Similar questions

I f s i n^{10} x - c o s^{10} x = 1 t h e n x = (n ϵ Z)

Q. The general solution of

s i n^{2} x - 2 c o s x + \frac{1}{4} = 0 i s x = (n ϵ Z)

Q. General solution of

x

for which

{cos}^{2} x = \frac{sin x tan x}{6}

is :

The general solution of $s i n x = s i n α, α ϵ [- \frac{π}{2}, \frac{π}{2}]$ is

The general solution of $t a n x = t a n α, α ϵ (- \frac{π}{2}, \frac{π}{2})$ is

Join BYJU'S Learning Program

The general solution to sin10x+cos10x=2916cos42x is

The general solution to $s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$ is