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Question: The value of \(x\), \[\\{ x \in R:\cos 2x + 2{\cos ^2}x = 2\\} \;\] would be....

The value of xx, x∈R:cos⁡2x+2cos⁡2x=2  \\{ x \in R:\cos 2x + 2{\cos ^2}x = 2\\} \; would be.

Explanation

Solution

In order to solve it we need to manipulate the given equation in such a way that we should get xx by itself. In order to get xx by itself we can perform any arithmetic operations on both LHS and RHS equally at the same time such that the equality of the given equation doesn’t change.
cos⁡(2x)=2 cos⁡2(x)−1\cos \left( {2x} \right) = 2{\text{ }}{\cos ^2}\left( x \right)-1
So by using the above formula and statement we can solve the above given question.

Complete step by step answer:
Given
x∈R:cos⁡2x+2cos⁡2x=2  ..........................(i)\\{ x \in R:\cos 2x + 2\cos 2x = 2\\} \;..........................\left( i \right)
Now we have to solve for xxin the above equation. It has been given that xxis an element ofRR.
Now let’s manipulate the given equation using trigonometric properties.
We know thatcos⁡(2x)=2 cos⁡2(x)−1\cos \left( {2x} \right) = 2{\text{ }}{\cos ^2}\left( x \right)-1.
So substituting it (i) we can write:
cos⁡2x+2cos⁡2x=2 2cos⁡2x−1+2cos⁡2x=2...................(ii)  \cos 2x + 2{\cos ^2}x = 2 \\\ 2{\cos ^2}x - 1 + 2{\cos ^2}x = 2...................\left( {ii} \right) \\\
Now let’s simplify equation (ii), we get:

2cos⁡2x−1+2cos⁡2x=2 4cos⁡2x−1=2 4cos⁡2x=2+1 4cos⁡2x=3.................(iii)  2{\cos ^2}x - 1 + 2{\cos ^2}x = 2 \\\ 4{\cos ^2}x - 1 = 2 \\\ 4{\cos ^2}x = 2 + 1 \\\ 4{\cos ^2}x = 3.................\left( {iii} \right) \\\

Now let’s divide both RHS and LHS of (iii) with 4.
So we get:

4cos⁡2x=3 cos⁡2x=34..................(iv)  4{\cos ^2}x = 3 \\\ {\cos ^2}x = \dfrac{3}{4}..................\left( {iv} \right) \\\

Now let’s take the root of the terms in RHS and LHS of (iv).
So we get:

cos⁡2x=34 cos⁡x=34 cos⁡x=±32..................(v)  {\cos ^2}x = \dfrac{3}{4} \\\ \cos x = \sqrt {\dfrac{3}{4}} \\\ \cos x = \pm \dfrac{{\sqrt 3 }}{2}..................\left( v \right) \\\

Now from (v) we can write that:
x=cos⁡−1(±32) x=nπ±π6:n∈Z....................(vi)  x = {\cos ^{ - 1}}\left( { \pm \dfrac{{\sqrt 3 }}{2}} \right) \\\ x = n\pi \pm \dfrac{\pi }{6}:n \in Z....................\left( {vi} \right) \\\
Therefore from (vi) we can write that for x∈R:cos⁡2x+2cos⁡2x=2  \\{ x \in R:\cos 2x + 2{\cos ^2}x = 2\\} \;on solving we get:
x=nπ±π6:n∈Zx = n\pi \pm \dfrac{\pi }{6}:n \in Z

Note: Some other equations needed for solving these types of problem are:

1 + {\tan ^2}x = {\sec ^2}x \\\ \begin{array}{*{20}{l}} {\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\\ {\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin }^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1} \end{array} \\\

Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.