The number of values of x in the interval [0, | vedclass

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Question

(D) Given the equation $2 \sin^2 x + 5 \sin x - 3 = 0$.
Let $u = \sin x$. Then the equation becomes $2u^2 + 5u - 3 = 0$.
Factoring the quadratic: $2

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$3$

Vedclass · Answer

(D) Given the equation $2 \sin^2 x + 5 \sin x - 3 = 0$.
Let $u = \sin x$. Then the equation becomes $2u^2 + 5u - 3 = 0$.
Factoring the quadratic: $2u^2 + 6u - u - 3 = 0 \implies 2u(u + 3) - 1(u + 3) = 0 \implies (2u - 1)(u + 3) = 0$.
This gives $u = \frac{1}{2}$ or $u = -3$.
Since the range of $\sin x$ is $[-1, 1]$,$u = -3$ is rejected.
Thus,$\sin x = \frac{1}{2}$.
In the interval $[0, 3\pi]$,we look for solutions for $\sin x = \frac{1}{2}$.
In $[0, 2\pi]$,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
In the interval $(2\pi, 3\pi]$,we add $2\pi$ to the first solution: $x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$.
So,the values are $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}$.
The total number of values is $3$.

The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2 \sin^2 x + 5 \sin x - 3 = 0$ is

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