JEE Mains · Maths · STD 11 - Trigonometrical equations
If \(0 \le x \le \pi \) and \({81^{{{\sin }^2}x}} + {81^{{{\cos }^2}x}} = 30\), then \(x =\)
- A \(\pi /6\)
- B \(\pi /2\)
- C \(\pi /4\)
- D \(3\pi /4\)
Answer & Solution
Correct Answer
(A) \(\pi /6\)
Step-by-step Solution
Detailed explanation
(a) We have, \({81^{{{\sin }^2}x}} + {81^{{{\cos }^2}x}} = 30\) Now check by options, put \(x = \frac{\pi }{6}\) then \({(81)^{{{\sin }^2}\pi /6}} + {(81)^{{{\cos }^2}\pi /6}} = 30\) ==> \({(81)^{1/4}} + {(81)^{3/4}} = 30\) Hence \((a)\) is the correct answer.
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