Question

Solve the equation 4cos2β−4cosβ=3 given that 0≤β<2π.​

Answer 1

Answer:

I can't understand your question

Step-by-step explanation:

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Answer 2

Answer: β=2π/3 and 4π/3

Step-by-step explanation:

First, rearrange the equation so that the right side is zero. This equation is a quadratic equation in cosβ and can be factored as

(2cosβ−3)(2cosβ+1)=0

Based on this equation, 2cosβ−3=0 or 2cosβ+1=0. These simplify to cosβ=32 and cosβ=−12.

The first equation has no solutions since −1≤cosβ≤1.

The second equation has solutions of the form β=2π3+2πn or β=4π3+2πn where n=0,±1,±2,....

By selecting only those solutions in the interval [0,2π), β=2π/3 and 4π/3.