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

Problem

2*sin2(x)+sin(x)−1=0

Solution

1. Identify the equation as a quadratic in terms of sin(x)

2. Substitute u=sin(x) to rewrite the equation as a standard quadratic equation.

2*u2+u−1=0

3. Factor the quadratic expression by finding two numbers that multiply to −2 and add to 1

(2*u−1)*(u+1)=0

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

2*u−1=0⇒u=1/2

u+1=0⇒u=−1

5. Back-substitute sin(x) for u to find the values of x

sin(x)=1/2

sin(x)=−1

6. Determine the general solutions for x based on the unit circle.

x=π/6+2*n*π

x=(5*π)/6+2*n*π

x=(3*π)/2+2*n*π

Final Answer

x=π/6+2*n*π,(5*π)/6+2*n*π,(3*π)/2+2*n*π

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