Solve for x 2sin(x)^2-3sin(x)-2=0

Problem

2*sin2(x)−3*sin(x)−2=0

Solution

1. Substitute a variable to simplify the quadratic form by letting u=sin(x)

2*u2−3*u−2=0

2. Factor the quadratic equation by finding two numbers that multiply to 2⋅(−2)=−4 and add to −3

(2*u+1)*(u−2)=0

3. Solve for u by setting each factor equal to zero.

2*u+1=0⇒u=−1/2

u−2=0⇒u=2

4. Back-substitute u=sin(x) into the equations.

sin(x)=−1/2

sin(x)=2

5. Evaluate the solutions for x Since the range of the sine function is [−1,1] the equation sin(x)=2 has no real solution.

sin(x)=2⇒No solution

6. Determine the general solutions for sin(x)=−1/2 using the unit circle. The reference angle is π/6 and sine is negative in the third and fourth quadrants.

x=(7*π)/6+2*n*π

x=(11*π)/6+2*n*π

Final Answer

x=(7*π)/6+2*n*π,(11*π)/6+2*n*π

Want more problems? Check here!