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Trigonometrical Equations
hard
समीकरण $1+\sin ^{4} x =\cos ^{2} 3 x , x \in\left[-\frac{5 \pi}{2}, \frac{5 \pi}{2}\right]$ के हलों की संख्या हैं
A
$3$
B
$4$
C
$5$
D
$7$
(JEE MAIN-2019)
Solution
$1\, + \,{\sin ^4}x\, = \,{\cos ^2}3x$
$\sin \,x\, = \,0$ and $\cos \,3x\, = 1$
$0,\,\,2\pi ,\, – \,2\pi ,\, – \,\pi ,\pi $
Standard 11
Mathematics