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

Problem

2*cos2(x)+3*sin(x)−3=0

Solution

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

2*(1−sin2(x))+3*sin(x)−3=0

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

2−2*sin2(x)+3*sin(x)−3=0

−2*sin2(x)+3*sin(x)−1=0

3. Multiply by -1 to make the leading coefficient positive, making it easier to factor.

2*sin2(x)−3*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 any integer.

sin(x)=1/2⇒x=π/6+2*n*π

sin(x)=1/2⇒x=(5*π)/6+2*n*π

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

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