Question:
The number of values of the triplet (a, b, c) for which $a \cos 2x + b \sin² x + c = 0$ is satisfied by all real x, is
The number of values of the triplet (a, b, c) for which $a \cos 2x + b \sin² x + c = 0$ is satisfied by all real x, is
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The number of values of the triplet (a, b, c) for which $a \cos 2x + b \sin x + c = 0$ is satisfied by all real x, is
Updated On: Apr 15, 2026
- 0
- 2
- 3
- infinite
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Use the identity $\cos 2x = 1 - 2\sin^2 x$.
Step 2: Analysis
Substituting into the equation: $a(1 - 2\sin^2 x) + b\sin^2 x + c = 0$, which simplifies to $(b - 2a)\sin^2 x + (a + c) = 0$.
Step 3: Evaluation
For this to be an identity satisfied for all $x$, the coefficients must be zero: $b - 2a = 0$ and $a + c = 0$.
Step 4: Conclusion
This implies $b = 2a$ and $c = -a$. Since '$a$' can be any real number, there are infinite triplets $(a, 2a, -a)$.
Final Answer: (d)
Use the identity $\cos 2x = 1 - 2\sin^2 x$.
Step 2: Analysis
Substituting into the equation: $a(1 - 2\sin^2 x) + b\sin^2 x + c = 0$, which simplifies to $(b - 2a)\sin^2 x + (a + c) = 0$.
Step 3: Evaluation
For this to be an identity satisfied for all $x$, the coefficients must be zero: $b - 2a = 0$ and $a + c = 0$.
Step 4: Conclusion
This implies $b = 2a$ and $c = -a$. Since '$a$' can be any real number, there are infinite triplets $(a, 2a, -a)$.
Final Answer: (d)
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