Question

solve
2 cos^2 x + 3 sin x = 0​

Answer 1

Step-by-step explanation:

$2[1−sin2x]+3sinx=0$

$2−2sin2x+3sinx=0$

$2sin </p><p>2</p><p> x−3sinx−2=0$

$Let sinx=y$

$2y </p><p>2</p><p> −3y−2=0$

$2y </p><p>2</p><p> −4y+y−2=0$

$2y(y−2)+1(y−2)=0$

$2y \:( +1)(y−2)=0$

$y \: = \frac{1}{2} ; \: 2$

$sinx=− </p><p>2</p><p>1</p><p> </p><p> ;sinx=2 \: not \: possible$

$sin \: is \: negative \: in \: third \: and \: \\ fourth \: quadrant</p><p>$

$</p><p></p><p>x=π+ </p><p>6</p><p>π</p><p> </p><p> ;2π− </p><p>6</p><p>π</p><p>$

$</p><p>x= </p><p>6</p><p>7π</p><p> </p><p> ; </p><p>6</p><p>11π</p><p> </p><p>$