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

7 \frac{π}{8}, 11 \frac{π}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9 \frac{π}{8}, 13 \frac{π}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11 \frac{π}{8}, 15 \frac{π}{8}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

15 \frac{π}{8}, 19 \frac{π}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $11 \frac{π}{8}, 15 \frac{π}{8}$
$sin 2 x + cos 2 x = 0$
$sin 2 x = - cos 2 x$
$\frac{sin 2 x}{cos 2 x} = - 1$
$tan 2 x = - 1 ⟶ 1$
$π < x < 2 π$
$2 π < 2 x < 4 π$
$tan x$ is negative in second and fourth quadrant
$∴ tan \frac{3 π}{4} = tan \frac{7 π}{4} = . . . . . = tan (4 n - 1) \frac{π}{4} = - 1$
$n = 1, 2, 3, . . . .$
But $2 π < 2 x < 4 π$
$∴ 2 x = \frac{11 π}{4}$ or $2 x = \frac{15 π}{4}$
$x = \frac{11 π}{8}$ or $\frac{15 π}{8}$

Suggest Corrections

Similar questions

sin 2 x + cos 2 x = 0

π < x < 2 π

\int [{sin}^{2} (\frac{9 π}{8} + \frac{x}{4}) - {sin}^{2} (\frac{7 π}{8} + \frac{x}{4})] d x

f (x) = cos (2 x + \frac{π}{4})

\frac{3 π}{8} < x < \frac{7 π}{8}

f (x) = sin 2 x + cos 2 x \forall x \in [0, \frac{π}{2}]

{cos}^{2} x + cos x = {sin}^{2} x

x \in [0, π]

Join BYJU'S Learning Program