Question

The statement
(3sin^x-2cosⓇx+y - 2 sinⓇx+ 3 cos* x)* =
= 9
is true for​

Answer 1

Step-by-step explanation:

We have,

$(3 \sin ^{4} x - 2 \cos^{6} x + y - 2 \sin^{6} x + 3 \cos^{4} x) ^{2} = 9$

$\implies [3 (\sin ^{4} x + \cos ^{4} x) - 2(\sin^{6} x + \cos^{6} x )+ y ] ^{2} = 9$

$\implies [3 \{(\sin ^{2} x + \cos ^{2} x)^{2} - 2 \sin^{2}x \cos^{2} x \}- 2 \{(\sin^{2} x + \cos^{2} x ) ( \sin ^{4} + \cos ^{4} x - \sin^{2} x \cos^{2}x) \}+ y ] ^{2} = 9$

$\implies [3 (1- 2 \sin^{2}x \cos^{2} x )- 2 \{(1)( (\sin ^{2} + \cos ^{2} x )^{2} - 2 \sin^{2}x \cos^{2}x - \sin^{2} x \cos^{2}x) \}+ y ] ^{2} = 9$

$\implies [3 (1- 2 \sin^{2}x \cos^{2} x )- 2 (1 - 3 \sin^{2}x \cos^{2}x ) + y ] ^{2} = 9$

$\implies [3- 6 \sin^{2}x \cos^{2} x - 2 + 6 \sin^{2}x \cos^{2}x + y ] ^{2} = 9$

$\implies (1+ y ) ^{2} = 9$

$\implies 1+ y^{2} + 2y = 9$

$\implies y^{2} + 2y - 8 = 0$

$\implies y^{2} + 4y - 2y- 8 = 0$

$\implies y(y + 4) - 2(y + 4) = 0$

$\implies (y - 2)(y + 4) = 0$

$\implies y = 2 \: \: or \: \: y = - 4$