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Question: How do you solve the expression \(2{{\cos }^{2}}x-5\cos x+3=0\) ?...

How do you solve the expression 2cos⁡2x−5cos⁡x+3=02{{\cos }^{2}}x-5\cos x+3=0 ?

Explanation

Solution

We have been given a trigonometric equation which consists of cosine function only. Since, the square of cosine of x is also present along with a constant term in the function, thus we shall treat it as a quadratic equation and hence factorise the middle term, cos x, into two terms which would be further grouped. Then we shall equate the linear equations thus formed to compute the final solution.

Complete step by step solution:
Given that, 2cos⁡2x−5cos⁡x+3=02{{\cos }^{2}}x-5\cos x+3=0
Since, the method of factoring the quadratic equation makes our calculations simpler, therefore, we use it the most.
For any quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0,
the sum of the roots =−ba=-\dfrac{b}{a} and the product of the roots =ca=\dfrac{c}{a}.
Thus, for the equation, 2cos⁡2x−5cos⁡x+3=02{{\cos }^{2}}x-5\cos x+3=0, a=2,a=2, b=−5b=-5 and c=3c=3.
We will find numbers by hit and trial whose product is equal to 2×3=62\times 3=6 and whose sum is equal to -5.
Such two numbers are -2 and -3 as −3+(−2)=−5-3+\left( -2 \right)=-5 and −3×−2=6-3\times -2=6.
Now, factoring the equation:
⇒2cos⁡2x−2cos⁡x−3cos⁡x+3=0\Rightarrow 2{{\cos }^{2}}x-2\cos x-3\cos x+3=0
Taking common, we get:
⇒2cos⁡x(cos⁡x−1)−3(cos⁡x−1)=0 ⇒(cos⁡x−1)(2cos⁡x−3)=0 \begin{aligned} & \Rightarrow 2\cos x\left( \cos x-1 \right)-3\left( \cos x-1 \right)=0 \\\ & \Rightarrow \left( \cos x-1 \right)\left( 2\cos x-3 \right)=0 \\\ \end{aligned}
For cos⁡x−1=0\cos x-1=0,
⇒cos⁡x=1\Rightarrow \cos x=1
This holds true for x=0x=0 in the interval 0≤x≤2π0\le x\le 2\pi .
For 2cos⁡x−3=02\cos x-3=0,
⇒cos⁡x=32\Rightarrow \cos x=\dfrac{3}{2}
But the range of cosines of x lies in the interval [−1,1]\left[ -1,1 \right] for the interval 0≤x≤2π0\le x\le 2\pi . Thus, we shall avoid this value of cos x.
Therefore, the solution of 2cos⁡2x−5cos⁡x+3=02{{\cos }^{2}}x-5\cos x+3=0 in the interval 0≤x≤2π0\le x\le 2\pi is x=0x=0 only.

Note: Another method solving the given equation was by substituting cos x as some variable-m in the beginning itself. Then, we could have obtained the entire equation as a quadratic equation in variable-m. Further, we could have also used the discriminant method of obtaining the roots of the equation whose values would be changed accordingly as we would re-substitute m=cos⁡x.m=\cos x.