If 3 cos x neq 2 sin x,then the general solut | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$ Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$: $\sin^2 x - (1 - 2\s

Vedclass · Accepted Answer

$(4n \pm 1) \frac{\pi}{2}, n \in \mathbb{Z}$

Vedclass · Answer

(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$ Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$: $\sin^2 x - (1 - 2\sin^2 x) = 2 - 2\sin x \cos x$ $3\sin^2 x - 1 = 2 - 2\sin x \cos x$ Alternatively,using $\sin^2 x = 1 - \cos^2 x$ and $\cos 2x = 2\cos^2 x - 1$: $(1 - \cos^2 x) - (2\cos^2 x - 1) = 2 - 2\sin x \cos x$ $2 - 3\cos^2 x = 2 - 2\sin x \cos x$ $-3\cos^2 x + 2\sin x \cos x = 0$ $\cos x (2\sin x - 3\cos x) = 0$ Since $3\cos x 
eq 2\sin x$,we have $2\sin x - 3\cos x 
eq 0$. Therefore,$\cos x = 0$. The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = (4n \pm 1) \frac{\pi}{2}$ for $n \in \mathbb{Z}$.

If $3 \cos x \neq 2 \sin x$,then the general solution of $\sin^2 x - \cos 2x = 2 - \sin 2x$ is

Similar Questions

If $f(x) = \frac{x^2}{x+a}$, then $f^{\prime \prime}(a)$ is equal to

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

Brush bordered epithelium is found in

The abscissae of $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are the roots of the equation $y^2+2py-q^2=0$. Find the equation of the circle with $AB$ as diameter.

Vedclass Test Series

Exam Paper Generator

Online Exam Module