The number of solutions of the equation 3cos^ | vedclass

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

Question

(B) Given the equation $3\cos^2x - 8\sin x = 0$. Using the identity $\cos^2x = 1 - \sin^2x$,we get $3(1 - \sin^2x) - 8\sin x = 0$. This simplifies to

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the equation $3\cos^2x - 8\sin x = 0$. Using the identity $\cos^2x = 1 - \sin^2x$,we get $3(1 - \sin^2x) - 8\sin x = 0$. This simplifies to $3 - 3\sin^2x - 8\sin x = 0$,or $3\sin^2x + 8\sin x - 3 = 0$. Let $t = \sin x$. Then $3t^2 + 8t - 3 = 0$. Factoring the quadratic: $3t^2 + 9t - t - 3 = 0 \implies 3t(t + 3) - 1(t + 3) = 0 \implies (3t - 1)(t + 3) = 0$. So,$t = \frac{1}{3}$ or $t = -3$. Since $-1 \leq \sin x \leq 1$,we discard $t = -3$. Thus,$\sin x = \frac{1}{3}$. In the interval $[0, 2\pi]$,there are $2$ solutions for $\sin x = \frac{1}{3}$. In the interval $[2\pi, 3\pi]$,$\sin x$ is positive,so there is $1$ additional solution. Total number of solutions in $[0, 3\pi]$ is $2 + 1 = 3$.

The number of solutions of the equation $3\cos^2x - 8\sin x = 0$ in the interval $[0, 3\pi]$ is

Similar Questions

For $n \in \mathbb{Z}$,the general solution of the trigonometric equation $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$ is

The smallest positive values of $x$ and $y$ which satisfy $\tan (x - y) = 1$ and $\sec (x + y) = \frac{2}{\sqrt{3}}$ are

The number of solution$(s)$ of the equation $\sin 2\theta + \cos 2\theta = -\frac{1}{2}$ for $\theta \in \left( 0, \frac{\pi}{2} \right)$ is:

The principal solutions of $\cot x = \sqrt{3}$ are

If $x \in(-\pi, \pi)$ then the number of solutions of the equation $2 \sin x \sin 3 x \sin 5 x+\sin 5 x \cos 4 x=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module