Question

Principal solutions of the equation \[\sin 2x + \cos 2x = 0\] , where \[\pi < x < 2\pi \] are
\[
  A.{\text{ }}\dfrac{{7\pi }}{8},\dfrac{{11\pi }}{8} \\
  B.{\text{ }}\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8} \\
  C.{\text{ }}\dfrac{{11\pi }}{8},\dfrac{{15\pi }}{8} \\
  D.{\text{ }}\dfrac{{15\pi }}{8},\dfrac{{19\pi }}{8} \\
 \]

Answer 1

Hint: In order to find out the principal solution, we will first find out the value of 2x, by using the value of tangent at different-different angles which are $\tan \dfrac{{3\pi }}{4} = \tan \dfrac{{7\pi }}{4} = \tan \dfrac{{11\pi }}{4} = \tan \dfrac{{15\pi }}{4} = - 1$. We will bring it in the form of tan to proceed.

Complete Step-by-Step solution:
Given that: \[\pi < x < 2\pi \]
Multiplying the inequality by 2 we get:
$ \Rightarrow 2\pi < 2x < 4\pi ............(1)$
Also given that \[\sin 2x + \cos 2x = 0\] ………. (2)
Finding the value of x from equation (2) by simplification
\[
   \Rightarrow \sin 2x + \cos 2x = 0 \\
   \Rightarrow \sin 2x = - \cos 2x \\
   \Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}} = - 1 \\
   \Rightarrow \tan 2x = - 1{\text{ }}\left[ {\because \dfrac{{\sin x}}{{\cos x}} = \tan x} \right] \\
   \Rightarrow 2x = {\tan ^{ - 1}}\left( { - 1} \right)..........(3) \\
 \]
Now we know the general form of tan function when it is -1.
$
   \Rightarrow \tan \dfrac{{3\pi }}{4} = \tan \dfrac{{7\pi }}{4} = \tan \dfrac{{11\pi }}{4} = \tan \dfrac{{15\pi }}{4} = - 1 \\
   \Rightarrow \tan \dfrac{{\left( {4n + 3} \right)\pi }}{4} = - 1...........(4) \\
 $
From comparing equation (3) and equation (4) we get:
The possible values of 2x are:
$\dfrac{{3\pi }}{4},\dfrac{{7\pi }}{4},\dfrac{{11\pi }}{4},\dfrac{{15\pi }}{4},...........,\dfrac{{\left( {4n + 3} \right)\pi }}{4}$
But from equation (1) we have the values of 2x lies between:
$2\pi < 2x < 4\pi $
So the values of 2x can be $\dfrac{{11\pi }}{4}{\text{ and }}\dfrac{{15\pi }}{4}$
So, let us divide the equation by 2 in order to get x:
$
  \because 2x = \dfrac{{11\pi }}{4},\dfrac{{15\pi }}{4} \\
   \Rightarrow x = \dfrac{{11\pi }}{{4 \times 2}},\dfrac{{15\pi }}{{4 \times 2}} \\
   \Rightarrow x = \dfrac{{11\pi }}{8},\dfrac{{15\pi }}{8} \\
 $
Hence, the principal solution of the equation is $x = \dfrac{{11\pi }}{8},\dfrac{{15\pi }}{8}$
So, option C is the correct option.

Note: A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. Solving an equation means to find the set of all values of the unknown value which satisfy the given equation. The solution lying between $0 - 2\pi $ is called the principal solution. But whenever the interval is mentioned in the problem students must find within that range.