Question

2 sin^2 (x) -5 sin (x) -3=0

Answer 1

we have that

$2sin^{2} x-5sin x-3=0$

I. Rewrite the equation by substituting the expression u in for sin x.

$2u^{2} -5u-3=0$

II. Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.

$2u^{2} -5u-3=0$ is equal to

using a graph calculator-----> see the attached figure

$(u-3)*(2u+1)=0$

III. Use the zero product property to solve the quadratic equation.

$(u-3)*(2u+1)=0$

(u-3)=0--------------> u=3

(2u+1)=0-------- 2u=-1------> u=-1/2-----> u=-0.5

IV. Rewrite your solutions to Part III by replacing u with sin x.

sin x=3--------> is not the solution (sin x can not be greater than 1)

sin x=-0.50------>is the solution

V. Solve the remaining equations for x, giving all solutions to the equation.

sin x=-0.50

if the sine is negative

then

x belong to the III or IV quadrant

we know that

sin 30°=0.50

so

the solution for the III quadrant is

x=180°+30°-------> x=210°

the solution for the IV quadrant is

x=360°-30°------> x=330°