Question

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

Answer 1

Hint: This question belongs to the topic of trigonometry. In this question, first we will put the value of \[\cos x\] 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 \[2{{\cos }^{2}}x-5\cos x=3\]. And, it is also said that the range of x is \[0\le x\le 2\pi \] that is in between \[0\] and \[2\pi \].
The equation which we have to solve is \[2{{\cos }^{2}}x-5\cos x=3\] can also be written as
\[\Rightarrow 2{{\cos }^{2}}x-5\cos x-3=0\]
To solve the above equation easily, we will put the value of \[\cos x\] as t.
Now, after putting the value of \[\cos x\] as t, we can write the above equation as
\[\Rightarrow 2{{t}^{2}}-5t-3=0\]
Now, we can factor the above equation easily.
The above equation can also be written as
\[\Rightarrow 2{{t}^{2}}-6t+t-3=0\]
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
\[\Rightarrow 2t\left( t-3 \right)+1\left( t-3 \right)=0\]
The above equation can also be written as
\[\Rightarrow \left( 2t+1 \right)\left( t-3 \right)=0\]
From here, we can say that
2t+1=0 and t-3=0
So, we can say from the above that
\[t=-\dfrac{1}{2}\]and \[t=3\]
As we have taken \[\cos x\] as t in the above equation. So, we can write
\[\cos x=-\dfrac{1}{2}\] and \[\cos x=3\]
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 \[\cos x\] cannot be 3. So, we can write only
\[\cos x=-\dfrac{1}{2}\]
As we know that \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\]. So, we can write the above equation as
\[\Rightarrow \cos x=-\cos \dfrac{\pi }{3}\]
The above equation can also be written as
\[\Rightarrow \cos x=\cos \left( \pi -\dfrac{\pi }{3} \right)=\cos \left( \dfrac{2\pi }{3} \right)\]
Solution for this trigonometric equation can be written as
\[x=2n\pi \pm \dfrac{2\pi }{3}\]
As the range of x is given as \[\left[ 0,2\pi \right]\], so by putting the value of n as 0 and 1 for the x to be in range. The value of x will be
\[x=\left( 0+\dfrac{2\pi }{3} \right)\] and \[x=\left( 2\pi -\dfrac{2\pi }{3} \right)\]
Or, we can write
\[x=\dfrac{2\pi }{3}\] and \[x=\dfrac{4\pi }{3}\]

Hence, the solutions for the equation \[2{{\cos }^{2}}x-5\cos x=3\] are \[\dfrac{2\pi }{3}\] and \[\dfrac{4\pi }{3}\].

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 \[\sin x\] and \[\cos x\]will be in the range of [-1,1].
Don’t forget the trigonometric values like at \[x=\dfrac{\pi }{3}\], \[\cos x=\dfrac{1}{2}\]. And, also don’t forget that \[\cos \left( \pi -x \right)=-\cos x\].
And, always remember the general solution for \[\cos x\]:
If \[\cos x=\cos \alpha \]
Then, general solution for x will be:
\[x=2n\pi \pm \alpha \], where n=0,1,2,3,........