Problem

Solve the equation 2cos2⁡x−3cos⁡x−2=0 for 0≤x≤2π

Answer

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Answer

Solving cos⁡x−2=0 gives cos⁡x=2. However, the range of cos⁡x is [−1,1], so this equation has no solutions.

Steps

Step 1 ：Rewrite the equation as a quadratic equation in terms of cos⁡x: 2cos2⁡x−3cos⁡x−2=0

Step 2 ：Factor the quadratic equation: (2cos⁡x+1)(cos⁡x−2)=0

Step 3 ：Set each factor equal to zero and solve for cos⁡x: 2cos⁡x+1=0 and cos⁡x−2=0

Step 4 ：Solving 2cos⁡x+1=0 gives cos⁡x=−12. The solutions to this equation in the interval [0,2π] are x=2π3 and x=4π3

Step 5 ：Solving cos⁡x−2=0 gives cos⁡x=2. However, the range of cos⁡x is [−1,1], so this equation has no solutions.

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