2sin2 -2 cos2x -1-cos- sin2x- x-2n-1-2- n- I- then cos2x is equal to

The correct option is B 3/5The given equation is equivalent to 2sin^2 ( ( π/2 )cos^2x )=2sin^2 ( ( π/2 )sin2x ) cos^2x=sin2x cos x(cos x 2sin x)=0 1 2tan ...

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

Byju's Answer

Standard XI

Mathematics

Trigonometric Equations

2sin2 π/2 co...

Question

$2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 1 - c o s (π s i n 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I$ , then $c o s 2 x$ is equal to

\frac{1}{5}

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

\frac{3}{5}

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

\frac{4}{5}

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

1

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

Open in App

Solution

The correct option is B $\frac{3}{5}$
The given equation is equivalent to
$2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 2 s i n^{2} ((\frac{π}{2}) s i n 2 x) c o s^{2} x = s i n 2 x c o s x (c o s x - 2 s i n x) = 0 1 - 2 t a n x = 0 a s c o s x \neq 0 s i n c e x \neq (2 n + 1) \frac{π}{2} t a n x = \frac{1}{2} \Rightarrow c o s 2 x = \frac{1 - t a n^{2} x}{1 + t a n^{2} x} = \frac{3}{5}$

Suggest Corrections

Similar questions

Q. If

2 sin^{2} ((\frac{π}{2}) cos^{2} x) = 1 - cos (π sin 2 x); x \neq \frac{(2 n + 1) π}{2}, n \in I

, then

cos 2 x

is equal to

Q. if

2 {sin}^{2} ((\frac{π}{2}) {cos}^{2} x) = 1 - cos (π {sin}^{2} 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

,then

cos 2 x

is equal to

Q. Show that

1 + {tan}^{2} x = {sec}^{2} x

where

x \neq (2 n + 1) \frac{π}{2}, n ϵ I

Q. If

A = {cos}^{2} x + \frac{1}{{cos}^{2} x}, B = cos x - \frac{1}{cos x} \forall x \neq (2 n \pm 1) \frac{π}{2}

, then the minimum value of

\frac{A}{B}

Q. The general solution of the equation

\frac{1 - sin x + . . . + {(- 1)}^{n} {sin}^{n} x + . . .}{1 + sin x + . . . + {sin}^{n} x} = \frac{1 - cos 2 x}{1 + cos 2 x}

x \neq (2 n + 1) \frac{π}{2}, n \in Z

Join BYJU'S Learning Program

2sin2((π2)cos2x)=1−cos(πsin2x),x≠(2n+1)π2,nϵI, then cos2x is equal to

The correct option is B 35The given equation is equivalent to 2sin2((π2)cos2x)=2sin2((π2)sin2x)cos2x=sin2xcos x(cos x−2sin x)=01−2tan x=0 as cosx≠0sincex≠(2n+1)π2tan x=12⇒cos2x=1−tan2x1+tan2x=35

$2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 1 - c o s (π s i n 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I$ , then $c o s 2 x$ is equal to