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Trigonometrical Equations
medium
If $4{\sin ^4}x + {\cos ^4}x = 1,$ then $x =$
A
$n\pi $
B
$n\pi \pm {\sin ^{ - 1}}\frac{2}{5}$
C
$n\pi + \frac{\pi }{6}$
D
None of these
Solution
(a) The given equation can be put in the form
$4{\sin ^4}x = 1 – {\cos ^4}x = (1 – {\cos ^2}x)\,(1 + {\cos ^2}x)$
$ \Rightarrow $ ${\sin ^2}x[4{\sin ^2}x – 1 – (1 – {\sin ^2}x)] = 0$
$ \Rightarrow $${\sin ^2}x[5{\sin ^2}x – 2] = 0$
$ \Rightarrow $$\sin x = 0$ or $\sin x = \pm \sqrt {2/5} $.
Hence $x = n\pi $ is the required answer.
Standard 11
Mathematics
