Solve for x 2sin(x)^2-9sin(x)+4=0

Problem

2*sin2(x)−9*sin(x)+4=0

Solution

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

2*u2−9*u+4=0

2. Factor the quadratic equation by finding two numbers that multiply to 2⋅4=8 and add to −9

(2*u−1)*(u−4)=0

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

2*u−1=0⇒u=1/2

u−4=0⇒u=4

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

sin(x)=1/2

sin(x)=4

5. Evaluate the solutions for x Since the range of the sine function is [−1,1] the equation sin(x)=4 has no real solutions.

sin(x)=4⇒No solution

6. Determine the general solutions for sin(x)=1/2 using the unit circle, where n is any integer.

x=π/6+2*n*π

x=(5*π)/6+2*n*π

Final Answer

x=π/6+2*n*π,(5*π)/6+2*n*π

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