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Trigonometrical Equations
hard
$x \in[0,2 \pi]$ की संख्या, जिनके लिए $\left|\sqrt{2 \sin ^{4} x+18 \cos ^{2} x}-\sqrt{2 \cos ^{4} x+18 \sin ^{2} x}\right|$ $=1$ है
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
