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

Problem

2*sin2(x)−cos(x)−1=0

Solution

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

2*(1−cos2(x))−cos(x)−1=0

2. Distribute and simplify the equation to form a quadratic equation in terms of cos(x)

2−2*cos2(x)−cos(x)−1=0

−2*cos2(x)−cos(x)+1=0

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

2*cos2(x)+cos(x)−1=0

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

(2*cos(x)−1)*(cos(x)+1)=0

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

2*cos(x)−1=0⇒cos(x)=1/2

cos(x)+1=0⇒cos(x)=−1

6. Solve for x using the unit circle values and general solutions, where n is any integer.

cos(x)=1/2⇒x=π/3+2*n*π,x=(5*π)/3+2*n*π

cos(x)=−1⇒x=π+2*n*π

Final Answer

x=π/3+2*n*π,(5*π)/3+2*n*π,π+2*n*π

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