If 2cos^2 x + 3sin x - 3 = 0 and 0^circ le x | vedclass

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

Question

(A) Given equation: $2\cos^2 x + 3\sin x - 3 = 0$Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:$2(1 - \sin^2 x) + 3\sin x - 3 = 0$$2 - 2\sin^2 x

Vedclass · Accepted Answer

$30^\circ, 90^\circ, 150^\circ$

Vedclass · Answer

(A) Given equation: $2\cos^2 x + 3\sin x - 3 = 0$Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:$2(1 - \sin^2 x) + 3\sin x - 3 = 0$$2 - 2\sin^2 x + 3\sin x - 3 = 0$$-2\sin^2 x + 3\sin x - 1 = 0$$2\sin^2 x - 3\sin x + 1 = 0$Factoring the quadratic equation:$(2\sin x - 1)(\sin x - 1) = 0$This gives two cases:$1) \sin x = \frac{1}{2} \Rightarrow x = 30^\circ, 150^\circ$$2) \sin x = 1 \Rightarrow x = 90^\circ$Thus,the values of $x$ are $30^\circ, 90^\circ, 150^\circ$.

If $2\cos^2 x + 3\sin x - 3 = 0$ and $0^\circ \le x \le 180^\circ$,then $x =$

Similar Questions

If $\theta$ and $\alpha$ are not odd multiples of $\frac{\pi}{2}$,then $\tan \theta = \tan \alpha$ implies the general solution is

If $0 \leq x \leq 2 \pi$,then the number of real values of $x$ which satisfy the equation $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$ is

If $\cos 2\theta = (\sqrt{2} + 1) \left( \cos \theta - \frac{1}{\sqrt{2}} \right)$,then the value of $\theta$ is

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

Find the general solution of $\sin x = -\frac{\sqrt{3}}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module