Question

sin^2 x - cos 2x = 2 - sin 2x if

A) x=n π/2, n ∈ l

B) tan x = 3/2

C) x = (2n+1) π/2, n ∈ l

D) x = n + π(-1)^n sin^-1 (2/3), n ∈ I

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sin^{2}(x)-cos(2x)=2-sin(2x)}$

$\sf{\implies\,sin^{2}(x)-\{cos^{2}(x)-sin^{2}(x)\}=2-2sin(x)cos(x)}$

$\sf{\implies\,sin^{2}(x)-cos^{2}(x)+sin^{2}(x)=2-2sin(x)cos(x)}$

$\sf{\implies\,2sin^{2}(x)-cos^{2}(x)=2-2sin(x)cos(x)}$

$\sf{\implies\,2sin(x)cos(x)-cos^{2}(x)=2-2sin^{2}(x)}$

$\sf{\implies\,2sin(x)cos(x)-cos^{2}(x)=2cos^{2}(x)}$

$\sf{\implies\,2sin(x)cos(x)=3cos^{2}(x)}$

$\sf{\implies\,2sin(x)cos(x)-3cos^{2}(x)=0}$

$\sf{\implies\,cos(x)\{2sin(x)-3cos(x)\}=0}$

$\sf{\implies\,cos(x)=0\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,2sin(x)-3cos(x)=0}$

$\sf{\implies\,x=(2n+1)\dfrac{\pi}{2}\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,tan(x)=\dfrac{3}{2}}$