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Question: How do you solve \[2{{\cos }^{2}}x-5\cos x=3\] in the interval\[0\le x\le 2\pi \]?...

How do you solve 2cos⁡2x−5cos⁡x=32{{\cos }^{2}}x-5\cos x=3 in the interval0≤x≤2π0\le x\le 2\pi ?

Explanation

Solution

This question belongs to the topic of trigonometry. In this question, first we will put the value of cos⁡x\cos x as t. After that, we will find the roots of the quadratic equation that will be seen in the solution.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to solve the equation 2cos⁡2x−5cos⁡x=32{{\cos }^{2}}x-5\cos x=3. And, it is also said that the range of x is 0≤x≤2π0\le x\le 2\pi that is in between 00 and 2π2\pi .
The equation which we have to solve is 2cos⁡2x−5cos⁡x=32{{\cos }^{2}}x-5\cos x=3 can also be written as
⇒2cos⁡2x−5cos⁡x−3=0\Rightarrow 2{{\cos }^{2}}x-5\cos x-3=0
To solve the above equation easily, we will put the value of cos⁡x\cos x as t.
Now, after putting the value of cos⁡x\cos x as t, we can write the above equation as
⇒2t2−5t−3=0\Rightarrow 2{{t}^{2}}-5t-3=0
Now, we can factor the above equation easily.
The above equation can also be written as
⇒2t2−6t+t−3=0\Rightarrow 2{{t}^{2}}-6t+t-3=0
Now, on the left side of the equation we can see that the first two are a factor of 2 and t. So, we will take common factors out 2 and t from the first two terms and 1 from the last two terms. Now, we can write the above equation as
⇒2t(t−3)+1(t−3)=0\Rightarrow 2t\left( t-3 \right)+1\left( t-3 \right)=0
The above equation can also be written as
⇒(2t+1)(t−3)=0\Rightarrow \left( 2t+1 \right)\left( t-3 \right)=0
From here, we can say that
2t+1=0 and t-3=0
So, we can say from the above that
t=−12t=-\dfrac{1}{2}and t=3t=3
As we have taken cos⁡x\cos x as t in the above equation. So, we can write
cos⁡x=−12\cos x=-\dfrac{1}{2} and cos⁡x=3\cos x=3
As we know that function of cos has always a range from -1 to +1 that [-1,1]
So, we can say that the value of cos⁡x\cos x cannot be 3. So, we can write only
cos⁡x=−12\cos x=-\dfrac{1}{2}
As we know that cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2}. So, we can write the above equation as
⇒cos⁡x=−cos⁡π3\Rightarrow \cos x=-\cos \dfrac{\pi }{3}
The above equation can also be written as
⇒cos⁡x=cos⁡(π−π3)=cos⁡(2π3)\Rightarrow \cos x=\cos \left( \pi -\dfrac{\pi }{3} \right)=\cos \left( \dfrac{2\pi }{3} \right)
Solution for this trigonometric equation can be written as
x=2nπ±2π3x=2n\pi \pm \dfrac{2\pi }{3}
As the range of x is given as [0,2π]\left[ 0,2\pi \right], so by putting the value of n as 0 and 1 for the x to be in range. The value of x will be
x=(0+2π3)x=\left( 0+\dfrac{2\pi }{3} \right) and x=(2π−2π3)x=\left( 2\pi -\dfrac{2\pi }{3} \right)
Or, we can write
x=2π3x=\dfrac{2\pi }{3} and x=4π3x=\dfrac{4\pi }{3}

Hence, the solutions for the equation 2cos⁡2x−5cos⁡x=32{{\cos }^{2}}x-5\cos x=3 are 2π3\dfrac{2\pi }{3} and 4π3\dfrac{4\pi }{3}.

Note: We should have a very deep knowledge in trigonometry to solve this type of question. And, also we should have a little bit of knowledge in quadratic equations, so that we can find the solutions of quadratic equations. Always remember that the value of sin⁡x\sin x and cos⁡x\cos xwill be in the range of [-1,1].
Don’t forget the trigonometric values like at x=π3x=\dfrac{\pi }{3}, cos⁡x=12\cos x=\dfrac{1}{2}. And, also don’t forget that cos⁡(π−x)=−cos⁡x\cos \left( \pi -x \right)=-\cos x.
And, always remember the general solution for cos⁡x\cos x:
If cos⁡x=cos⁡α\cos x=\cos \alpha
Then, general solution for x will be:
x=2nπ±αx=2n\pi \pm \alpha , where n=0,1,2,3,........