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

Problem

2*cos2(x)+sin(x)−1=0

Solution

1. Apply the Pythagorean identity to rewrite the equation in terms of a single trigonometric function using cos2(x)=1−sin2(x)

2*(1−sin2(x))+sin(x)−1=0

2. Distribute and simplify the terms to form a quadratic equation.

2−2*sin2(x)+sin(x)−1=0

−2*sin2(x)+sin(x)+1=0

3. Multiply by -1 to make the leading coefficient positive for easier factoring.

2*sin2(x)−sin(x)−1=0

4. Factor the quadratic expression by treating sin(x) as the variable.

(2*sin(x)+1)*(sin(x)−1)=0

5. Set each factor to zero to find the possible values for sin(x)

2*sin(x)+1=0⇒sin(x)=−1/2

sin(x)−1=0⇒sin(x)=1

6. Solve for x using the unit circle values and general solutions where n is an integer.
   For sin(x)=1

x=π/2+2*n*π

For sin(x)=−1/2

x=(7*π)/6+2*n*π

x=(11*π)/6+2*n*π

Final Answer

x=π/2+2*n*π,(7*π)/6+2*n*π,(11*π)/6+2*n*π

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