JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

The equation

{\sin ^4}x - 2{\cos ^2}x + {a^2} = 0

is solvable for

View answer and explanation

Correct option: B

- \,\sqrt 2 \le a \le \sqrt 2

Option B (

- \sqrt 2 \le a \le \sqrt 2

) is correct because it defines the range of 'a' for which the given equation has real solutions.

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