The number of x ∈ [0, 2 π] for which | √2 sin4 x + 18 cos2 x - √2 cos4 x + 18 sin2 x| = 1 is :

Question

The number of x ∈ [0, 2 π] for which | √2 sin4 x + 18 cos2 x - √2 cos4 x + 18 sin2 x| = 1 is : (A) 2 (B) 4 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

√2 sin4 x+18 cos2 x-√2 cos4 x+18 sin2 x=1 √2 sin4 x+18 cos2 x-√2 cos4 x+18 sin2 x=±1 √2 sin4 x+18 cos2 x=±1+√2 cos4 x+18 sin2 x by squaring both the sides we will get B solutions

Q. The number of $x \in [0, 2 π]$ for which $∣ 2 sin^{4} x + 18 cos^{2} x - 2 cos^{4} x + 18 sin^{2} x ∣ = 1$ is :

$2$

$4$

$6$

$8$

$2 s i n^{4} x + 18 co s^{2} x - 2 co s^{4} x + 18 s i n^{2} x = 1$
$2 s i n^{4} x + 18 co s^{2} x - 2 co s^{4} x + 18 s i n^{2} x = \pm 1$
$2 s i n^{4} x + 18 co s^{2} x = \pm 1 + 2 co s^{4} x + 18 s i n^{2} x$
by squaring both the sides we will get B solutions

Q. The number of x∈[0,2π] for which ∣2sin4x+18cos2x​−2cos4x+18sin2x​∣=1 is :

2

4

6

8

2sin4x+18cos2x​−2cos4x+18sin2x​=1 2sin4x+18cos2x​−2cos4x+18sin2x​=±1 2sin4x+18cos2x​=±1+2cos4x+18sin2x​ by squaring both the sides we will get B solutions

Q. The number of $x \in [0, 2 π]$ for which $∣ 2 sin^{4} x + 18 cos^{2} x - 2 cos^{4} x + 18 sin^{2} x ∣ = 1$ is :

$2$

$4$

$6$

$8$

$2 s i n^{4} x + 18 co s^{2} x - 2 co s^{4} x + 18 s i n^{2} x = 1$
$2 s i n^{4} x + 18 co s^{2} x - 2 co s^{4} x + 18 s i n^{2} x = \pm 1$
$2 s i n^{4} x + 18 co s^{2} x = \pm 1 + 2 co s^{4} x + 18 s i n^{2} x$
by squaring both the sides we will get B solutions