Question

Solve the following trigonometric question $ {\sin ^6}2x + {\cos ^6}2x = \dfrac{7}{{16}} $ .

Answer 1

Hint: The given question is related to the concept of trigonometric equations. Trigonometric equation is an equation which involves one or more trigonometric ratios of an unknown angle. We know that the value of x remains between $ {0^ \circ } $ and $ {360^ \circ } $ . The domain of both $ \sin x $ and $ \cos x $ is $ \left( { - \infty ,\infty } \right) $ and range is between $ \left[ { - 1,1} \right] $ . Here, in this question we have to solve the given trigonometric equation in order to find the value of x. We will use trigonometric identity to solve the given problem.

Complete step by step solution:
Given is $ {\sin ^6}2x + {\cos ^6}2x = \dfrac{7}{{16}} $
We know the following trigonometric identity,
$ {\sin ^6}x + {\cos ^6}x = 1 - 3\left( {{{\sin }^2}x} \right)\left( {{{\cos }^2}x} \right) $
Using the same trigonometric identity in the given trigonometric question, we get,
$
   \Rightarrow {\left( 1 \right)^3} - 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{7}{{16}} \\
   \Rightarrow 1 - 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{7}{{16}} \\
   \Rightarrow - 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{7}{{16}} - 1 \\
   \Rightarrow - 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{{7 - 16}}{{16}} \\
   \Rightarrow - 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = - \dfrac{9}{{16}} \\
$
Now, since minus (-) sign is on both the sides of the equation, it automatically cancels off.
$
   \Rightarrow 3\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{9}{{16}} \\
   \Rightarrow 4\left( {{{\sin }^2}2x} \right)\left( {{{\cos }^2}2x} \right) = \dfrac{3}{4} \\
   \Rightarrow {\sin ^2}4x = \dfrac{3}{4} \\
$
Therefore,
$
  \therefore \sin 4x = \pm \sqrt {\dfrac{3}{4}} = \pm \dfrac{{\sqrt 3 }}{2} = \pm \sin \dfrac{\pi }{3} \\
  \therefore x = \dfrac{{n\pi }}{4} \pm \dfrac{\pi }{{12}} \\
$
Hence, $ x = \dfrac{{n\pi }}{4} \pm \dfrac{\pi }{{12}} $

Note: Here, in this question we used trigonometric identity and we could easily find the value. It is highly recommended to keep all the trigonometric identities in mind while solving trigonometry questions as without using these identities, solving a question would be difficult and take a lot of time. Trigonometric identities are satisfied for every value of the unknown angle while the trigonometric equations are satisfied only for some values of unknown angle. Students should remember that the answer should not contain any such values of angles which might make any of the terms undefined and domain should not be changed. Students can find a lot of similar questions in their NCERT textbook. They should practice them for more clarity.