Solve for x 2sin(x)^2-3sin(x)+1=0

Problem

2*sin(x)−3*sin(x)+1=0

Solution

1. Substitute a variable to simplify the equation. Let u=sin(x)

2*u2−3*u+1=0

2. Factor the quadratic equation. We look for two numbers that multiply to 2 and add to −3

(2*u−1)*(u−1)=0

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

2*u−1=0⇒u=1/2

u−1=0⇒u=1

4. Back-substitute u=sin(x) to find the values of x

sin(x)=1/2

sin(x)=1

5. Determine the general solutions for x using the unit circle.
   For sin(x)=1/2

x=π/6+2*n*π

x=(5*π)/6+2*n*π

For sin(x)=1

x=π/2+2*n*π

Final Answer

x=π/6+2*n*π,(5*π)/6+2*n*π,π/2+2*n*π

Want more problems? Check here!