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Question: How do you solve the following equation \[4{\cos ^2}x - 3 = 0\] in the interval \[\left[ {0,2 \pi} \...

How do you solve the following equation 4cos⁡2x−3=04{\cos ^2}x - 3 = 0 in the interval [0,2π]\left[ {0,2 \pi} \right]?

Explanation

Solution

In the given question, we have been given a quadratic, binomial equation. We have to solve for the value of the given variable. This variable here is in the form of a trigonometric function. We solve it by taking the other term to the other side (its sign changes), taking the square root on both sides, then using the standard value table to find the value of the angle.

Complete step by step answer:
We have to solve 4cos⁡2x−3=04{\cos ^2}x - 3 = 0.
Taking the constants 33 and 44 to the other side,
cos⁡2x=34{\cos ^2}x = \dfrac{3}{4}
Taking square root on both sides,
cos⁡x=±32\cos x = \pm \dfrac{{\sqrt 3 }}{2}
Now, we know, cos⁡(π6)=32\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}
We also know that: cos⁡(2π−θ)=cos⁡θ\cos \left( {2\pi - \theta } \right) = \cos \theta
Hence, cos⁡(2π−π6)=32\cos \left( {2\pi - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}
Also, cos⁡(π±θ)=−cos⁡θ\cos \left( {\pi \pm \theta } \right) = - \cos \theta
Hence, cos⁡(π±π6)=−cos⁡(π6)\cos \left( {\pi \pm \dfrac{\pi }{6}} \right) = - \cos \left( {\dfrac{\pi }{6}} \right)
Thus, cos⁡(5π6)=cos⁡(7π6)=−32\cos \left( {\dfrac{{5\pi }}{6}} \right) = \cos \left( {\dfrac{{7\pi }}{6}} \right) = - \dfrac{{\sqrt 3 }}{2}

So, x=π6,5π6,7π6,11π6x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}

Note:
In the given question, we were given a quadratic equation which was a binomial – that is, it has two terms. We solved it by keeping the variable term on one side and the remaining term on the other side. Then we took the square root so that we can have the primitive form of the variable. Then we solved the argument of the function and got our answer.