Solve for x (tan(x)^2-9)(2cos(x)+1)=0

Problem

(tan2(x)−9)*(2*cos(x)+1)=0

Solution

1. Apply the zero product property by setting each factor equal to zero.

tan2(x)−9=0

2*cos(x)+1=0

2. Solve the first equation for tan(x) by adding 9 to both sides and taking the square root.

tan2(x)=9

tan(x)=±3

3. Find the general solution for x when tan(x)=±3 using the inverse tangent function.

x=arctan(3)+n*π

x=arctan(−3)+n*π

4. Solve the second equation for cos(x) by subtracting 1 and dividing by 2

2*cos(x)=−1

cos(x)=−1/2

5. Identify the angles where the cosine is −1/2 within the unit circle.

x=(2*π)/3+2*n*π

x=(4*π)/3+2*n*π

6. Combine all solutions into a single set where n is any integer.

x=±arctan(3)+n*π

x=(2*π)/3+2*n*π

x=(4*π)/3+2*n*π

Final Answer

x={±arctan(3)+n*π,(2*π)/3+2*n*π,(4*π)/3+2*n*π}

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