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Principal solutions of the equation, $\sin 2 x + \cos 2 x = 0$ where $π < x < 2 π$ are;

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\frac{7 π}{8}, \frac{11 π}{8}

\frac{9 π}{8}, \frac{13 π}{8}

\frac{11 π}{8}, \frac{15 π}{8}

\frac{15 π}{8}, \frac{19 π}{8}

answer is C.

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Detailed Solution

Given equation is,

\sin 2 x + \cos 2 x = 0

....(1)
It can be written as,

\begin{matrix} \Rightarrow \sin 2 x = - \cos 2 x \\ \Rightarrow \frac{\sin 2 x}{\cos 2 x} = - 1 \\ \Rightarrow \tan 2 x = - 1 \\ \Rightarrow 2 x = \tan^{- 1} (- 1) \dots \dots (2) \end{matrix}

We have given that,

π < x < 2 π

It can be written as,

2 π < 2 x < 4 π

....(3)
Now, we know that Since,

\tan \frac{3 π}{4} = \tan \frac{7 π}{4} = \tan \frac{11 π}{4} = \dots \dots

\Rightarrow \tan \frac{(4 n + 3) π}{4} = - 1

Here, the values of

2 x

can be

\frac{3 π}{4}, \frac{7 π}{4}, \frac{11 π}{4}, \dots ., \frac{(4 n + 3) π}{4}

Now from the equation 3, the possible values of

2 x

can only be

\frac{11 π}{4}

and

\frac{15 π}{4}

.
Thus,

\begin{matrix} 2 x = \frac{11 π}{4}, \frac{15 π}{4} \\ \Rightarrow x = \frac{11 π}{8}, \frac{15 π}{8} \end{matrix}

So, the required solution

for given equation are

\frac{11 π}{8}

and

\frac{15 π}{8}

.
Correct option is 3.

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