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

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

Explanation

Solution

We will convert the given trigonometric equation to a quadratic equation by equating cos⁡x\cos x with a variable c.c. Then we will solve the quadratic equation for cc using the quadratic formula.

Complete step by step solution:
Let us consider the given trigonometric equation 2cos⁡2x−5cos⁡x+2=0.2{{\cos }^{2}}x-5\cos x+2=0.
We are asked to solve the given trigonometric equation.
Let us first convert the given trigonometric function into a quadratic equation in cc by equating cos⁡x\cos x into c.c.
The trigonometric equation will become a quadratic equation 2c2−5c+2=0.2{{c}^{2}}-5c+2=0.
We can use the quadratic formula to solve the obtained quadratic equation.
Suppose that we have a quadratic equation ax2+bx+c=0.a{{x}^{2}}+bx+c=0. Then we can use the following formula which is called the quadratic formula to solve the equation for x:x=−b±b2−4ac2a.x: x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.
Let us compare the coefficients to get a=2,b=−5a=2, b=-5 and c=2.c=2.
So, the quadratic formula for our equation is c=−(−5)±(−5)2−4×2×22×2.c=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 2\times 2}}{2\times 2}.
We can write this equation as c=5±25−164.c=\dfrac{5\pm \sqrt{25-16}}{4}.
That can be written as c=5±94.c=\dfrac{5\pm \sqrt{9}}{4}.
When we write this equation by substituting 3=9,3=\sqrt{9}, we will get c=5±34.c=\dfrac{5\pm 3}{4}.
From this, we will get the possible values of the variable cc as c=5−34c=\dfrac{5-3}{4} or c=5+34.c=\dfrac{5+3}{4}.
And this will give us c=24=12c=\dfrac{2}{4}=\dfrac{1}{2} or c=84=2.c=\dfrac{8}{4}=2.
We know that cos⁡x=c.\cos x=c. So, we will get cos⁡x=2\cos x=2 or cos⁡x=12.\cos x=\dfrac{1}{2}.
But we know that the range of the Cosine function is [−1,1].\left[ -1,1 \right].
So, it is not possible that cos⁡x=2.\cos x=2.
Therefore, we will get cos⁡x=12.\cos x=\dfrac{1}{2}.
We know that the values of xx for which cos⁡x=12\cos x=\dfrac{1}{2} are ±π3.\pm \dfrac{\pi }{3}.
In general, x=±π3+2nπ,n∈Z.x=\pm \dfrac{\pi }{3}+2n\pi ,n\in \mathbb{Z}.

Hence the solution of the given equation is given by x=±π3+2nπ,n∈Z.x=\pm \dfrac{\pi }{3}+2n\pi ,n\in \mathbb{Z}.

Note: We can factorize the obtained quadratic equation 2c2−5c+2=02{{c}^{2}}-5c+2=0 as (c−2)(2c−1)=0.\left( c-2 \right)\left( 2c-1 \right)=0. Then from the first factor, we will get c=2c=2 and from the second fraction, we will get 2c=1⇒c=12.2c=1\Rightarrow c=\dfrac{1}{2}.