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Trigonometrical Equations
hard
$\left| {\sqrt {2\,{{\sin }^4}\,x\, + \,18\,{{\cos }^2}\,x} - \,\sqrt {2\,{{\cos }^4}\,x\, + \,18\,{{\sin }^2}\,x} } \right| = 1$ ના $x \in [0,2\pi ]$ માં ઉકેલોની સંખ્યા .......... છે.
A
$2$
B
$6$
C
$4$
D
$8$
(JEE MAIN-2016)
Solution
$|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}|=1$
$\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2} \cos ^{4} x+18 \sin ^{2} x=\pm 1$
$\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}=\pm 1+\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}$
by squaring both the sides we will get $8$ solutions
Standard 11
Mathematics
