Question

Solve:- 2cos²x+3sinx=0

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Answer 1

2 Cos[2]x + 3 Sinx = 0
2 ( 1 - Sin[2]x ) + 3 Sinx = 0
2 Sin[2]x - 3 Sinx - 2 = 0
2 Sin[2]x - 4 Sinx + Sinx - 2 = 0
( 2 Sinx + 1 ) ( Sinx - 2 ) = 0

Sinx = -1/2 = Sin 120 => x = 120

Sinx = 2

Answer 2

SOLUTION

$= > 2 \: cos {}^{2} x + 3 \: sin \: x = 0 \: \: \: (sin {}^{2}x + cos {}^{2} x = 1 = > cos {}^{2} x = 1 - sin {}^{2}x \\ = > 2( 1 - sin {}^{2} x) + 3 \: sin \: x = 0 \\ = > 2 - 2sin {}^{2} x + 3sinx = 0 \\ = > - 2sin {}^{2} x + 3sin \: x + 2 = 0 \\ let \: sin \: x = a \\ = > - 2 {a}^{2} + 3a + 2 = 0 \\ = > 0 = 2 {a}^{2} - 3a - 2 \\ = > 2 {a}^{2} - 3a - 2 = 0 \\ = > 2 {a}^{2} - 4a + a - 2 = 0 \\ = > 2a(a - 2) + 1(a - 2) = 0 \\ = > (2a + 1) (a - 2) = 0 \\ = > 2a + 1 = 0 \: \: \: \: \: \: or \: \: \: \: \: a - 2 = 0 \\ = > 2a = - 1 \: \: \: or \: \: \: a = 2 \\ = > a = - \frac{1}{2} \: \: and \: a \: = 2 \\ = > hence \: sin \: x = - \frac{1}{2} \: or \: \:sin \:x = 2$

Value of sin is always between-1 and 1

Hence sin x= 2 is not possible

Therefore

=) sin x= -1/2

hope it helps ✔️