Question:
The number of solutions of the trigonometric equation $\sin^2 x - \sin x - 2 = 0$ in the interval $[0, 2\pi]$ is:
The number of solutions of the trigonometric equation $\sin^2 x - \sin x - 2 = 0$ in the interval $[0, 2\pi]$ is:
Show Hint
Always check the range constraints of trigonometric functions first. Since $-1 \le \sin x \le 1$, roots outside this interval are instantly discarded.
Updated On: May 31, 2026
- $1$
- $2$
- $3$
- $0$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
This is a quadratic equation in terms of $\sin x$. We can solve for the roots of $\sin x$ using factorization and check which roots are valid within the range $[-1, 1]$.
Step 2: Meaning
Let $t = \sin x$. The equation becomes $t^2 - t - 2 = 0$.
Step 3: Analysis
Factorizing the quadratic equation: \[ (t - 2)(t + 1) = 0 \implies t = 2 \text{ or } t = -1 \] Since $t = \sin x$, we analyze both cases:
• Case 1: $\sin x = 2$. Since the range of the sine function is $[-1, 1]$, this equation has no real solutions.
• Case 2: $\sin x = -1$. In the interval $[0, 2\pi]$, this occurs only at $x = \frac{3\pi}{2}$. Thus, there is only $1$ valid solution in the given interval.
Step 4: Conclusion
The number of solutions of the trigonometric equation in $[0, 2\pi]$ is $1$. Final Answer: (A)
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