Question

The general solution of the equation $ 2\cos 2x=3\left( 2{{\cos }^{2}}x \right)-4 $ is,
\[\begin{align}
  & A.x=2n\pi ,n\in I \\
 & B.x=n\pi ,n\in I \\
 & C.x=\dfrac{n\pi }{4},n\in I \\
 & D.x=\dfrac{n\pi }{2},n\in I \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the general solution of the given equation $ 2\cos 2x=3\left( 2{{\cos }^{2}}x \right)-4 $ . For this, we will first change $ 2{{\cos }^{2}}x $ into the form of cos2x using the formula: $ 2{{\cos }^{2}}x=1-\cos 2x $ . After that, we will simplify the equation and find the value of cos2x. Then we will use the general solution of $ \cos x=\cos \alpha $ to find the value of x. For $ \cos x=\cos \alpha ,x=n\pi \pm \alpha $ . We will also use the value of $ \cos {{0}^{\circ }} $ from the trigonometric ratio table which is equal to 1.

Complete step by step answer:
Here we are given the equation as, $ 2\cos 2x=3\left( 2{{\cos }^{2}}x \right)-4 $ .
To find the value of x, we need to change the different trigonometric functions into a single trigonometric function. As we can see, $ 2{{\cos }^{2}}x $ can be converted in the form of cos2x. So applying the formula $ 2{{\cos }^{2}}x=1-\cos 2x $ in the given equation, we get, \[2\cos 2x=3\left( 1-\cos 2x \right)-4\].
Opening bracket we get, \[2\cos 2x=3-3\cos 2x-4\].
Taking a variable on one side and constant to the other side we get, \[2\cos 2x-3\cos 2x=3-4\Rightarrow -\cos 2x=-1\].
Canceling negative signs from both sides of the equation we get, \[\cos 2x=1\].
Now from the trigonometric ratio table, we know that $ \cos {{0}^{\circ }}=1 $ so we can write 1 as $ \cos {{0}^{\circ }} $ . Hence we get, $ \cos 2x=\cos {{0}^{\circ }} $ .
As we know, $ \cos x=\cos \alpha $ implies that $ \theta =2n\pi \pm \alpha $ where $ n\in I $ so here $ \theta =2x\text{ and }\alpha =0 $ . Applying this formula, we get,
 $ 2x=2n\pi \pm 0 $ where $ n\in I $ , $ 2x=2n\pi $ where $ n\in I $ .
Dividing both sides by 2, we get, $ x=n\pi $ where $ n\in I $ .
Hence the general solution of x is $ n\pi ,n\in I $ .
Hence option B is the correct answer.

Note:
Students should take care of the signs while solving these sums. Keep in mind all the general values of the trigonometric function. Students should note that here I represent the set of integers and n belong to this set which means that for any integer n, $ x=n\pi $ will satisfy the given equation.