Convert to quadratic in tan

11

1. Solve the equation \(3 \tan ^ { 2 } x - 5 \tan x - 2 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
2. Find the set of values of \(k\) for which the equation \(3 \tan ^ { 2 } x - 5 \tan x + k = 0\) has no solutions.
3. For the equation \(3 \tan ^ { 2 } x - 5 \tan x + k = 0\), state the value of \(k\) for which there are three solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\), and find these solutions.

4

1. Show that the equation $$\sin x - \cos x = \frac { 6 \cos x } { \tan x }$$ can be expressed in the form $$\tan ^ { 2 } x - \tan x - 6 = 0 .$$
2. Hence solve the equation \(\sin x - \cos x = \frac { 6 \cos x } { \tan x }\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

6 In this question you must show detailed reasoning. Solve the equation \(\tan x - 3 \cot x = 2\) for values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

Solve the equation $$6 \sec ^ { 2 } x - 8 = \tan x$$ for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

Solve the equation \(\tan(\theta - 60°) = 3 \cot \theta\) for \(-90° < \theta < 90°\). [5]

By first expressing the equation \(\tan(x + 45°) = 2 \cot x + 1\) as a quadratic equation in \(\tan x\), solve the equation for \(0° < x < 180°\). [6]

Solve the equation $$\tan(45° - x) = 2\tan x,$$ giving all solutions in the interval \(0° < x < 180°\). [5]

Solve, for \(0° < \theta < 360°\), the equation \(\sec^2 \theta = 4 \tan \theta - 2\). [5]

Find the values of \(x\) in the interval \(-180 < x < 180\) for which $$\tan (x + 45)^{\circ} - \tan x^{\circ} = 4,$$ giving your answers to 1 decimal place. [9]

Find all values of \(\theta\) in the interval \(-180 < \theta < 180\) for which $$\tan^2 \theta^\circ + \sec \theta^\circ = 1.$$ [6]

Solve the equation \(\cosec^2 \theta = 1 + 2 \cot \theta\), for \(-180° \leqslant \theta \leqslant 180°\). [6]

solve, for \(0° < \theta < 360°\), the equation $$2 \tan^2 \theta - \frac{1}{\cos \theta} = 4.$$ [4]

In this question you must show detailed reasoning. Solve the following equation for \(x\) in the interval \(0° < x < 180°\) $$1 + \log_3\left(1 + \tan^2 2x\right) = 2\log_3(-4\sin 2x)$$ [8]