Solve the equation ${\sec ^2}x + 2\tan x = 0,{\text{ }}0 \leqslant x \leqslant 2\pi$.

METHOD 1

use of ${\sec ^2}x = {\tan ^2}x + 1$     M1

${\tan ^2}x + 2\tan x + 1 = 0$

${(\tan x + 1)^2} = 0$     (M1)

$\tan x =  - 1$     A1

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$     A1A1

METHOD 2

$\frac{1}{{{{\cos }^2}x}} + \frac{{2\sin x}}{{\cos x}} = 0$     M1

$1 + 2\sin x\cos x = 0$

$\sin 2x =  - 1$     M1A1

$2x = \frac{{3\pi }}{2},{\text{ }}\frac{{7\pi }}{2}$

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$     A1A1

Note:     Award A1A0 if extra solutions given or if solutions given in degrees (or both).

[5 marks]