Q87 / 150 · MHT CET 2016 · Mathematics

The solution of the equation $\sin 2 x + \cos 2 x = 0,$ where $π < x < 2 π$ are

A
$\frac{7 π}{8}, \frac{11 π}{8}$
B
$\frac{9 π}{8}, \frac{13 π}{8}$
Correct answer
C
$\frac{11 π}{8}, \frac{15 π}{8}$
D
$\frac{15 π}{8}, \frac{19 π}{8}$

Solution

Given,

\sin 2 x + \cos 2 x = 0

.
Multiplying by

\frac{1}{\sqrt{2}}

on both side,

∴ \frac{1}{\sqrt{2}} \sin 2 x + \frac{1}{\sqrt{2}} \cos 2 x = 0

\Rightarrow \sin (\frac{π}{4} + 2 x) = 0

\Rightarrow 2 x + \frac{π}{4} = n π

\Rightarrow x = \frac{(4 n - 1) π}{8}

∴ x = \frac{11 π}{8} a n d \frac{15 π}{8}

Hence, the correct answer is option: 'C'.

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