To solve \(\tan 2x = 5 \cot x\), we start by using the identity \(\cot x = \frac{1}{\tan x}\).

Thus, the equation becomes:

\(\tan 2x = \frac{5}{\tan x}\)

Using the double angle formula for tangent, \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), we substitute:

\(\frac{2 \tan x}{1 - \tan^2 x} = \frac{5}{\tan x}\)

Cross-multiplying gives:

\(2 \tan^2 x = 5(1 - \tan^2 x)\)

Expanding and rearranging terms, we get:

\(2 \tan^2 x = 5 - 5 \tan^2 x\)

\(7 \tan^2 x = 5\)

\(\tan^2 x = \frac{5}{7}\)

Taking the square root, we find:

\(\tan x = \pm \sqrt{\frac{5}{7}}\)

Now, solve for \(x\) in the interval \(0^\circ < x < 180^\circ\).

For \(\tan x = \sqrt{\frac{5}{7}}\), we find:

\(x = \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 40.2^\circ\)

For \(\tan x = -\sqrt{\frac{5}{7}}\), we find:

\(x = 180^\circ - \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 139.8^\circ\)

Thus, the solutions are \(x = 40.2^\circ\) and \(x = 139.8^\circ\).

To solve the equation \(2 \cos x - \cos \frac{1}{2}x = 1\), we can use trigonometric identities and algebraic manipulation.

1. Use the double angle formula: \(\cos x = 2 \cos^2 \frac{x}{2} - 1\).

2. Substitute \(\cos x\) in the equation:

\(2(2 \cos^2 \frac{x}{2} - 1) - \cos \frac{x}{2} = 1\)

3. Simplify the equation:

\(4 \cos^2 \frac{x}{2} - 2 - \cos \frac{x}{2} = 1\)

4. Rearrange to form a quadratic equation in \(\cos \frac{x}{2}\):

\(4 \cos^2 \frac{x}{2} - \cos \frac{x}{2} - 3 = 0\)

5. Let \(u = \cos \frac{x}{2}\). The equation becomes:

\(4u^2 - u - 3 = 0\)

6. Solve the quadratic equation using the quadratic formula:

\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 4\), \(b = -1\), \(c = -3\).

\(u = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}\)

7. The solutions for \(u\) are:

\(u = 1\) and \(u = -\frac{3}{4}\).

8. Solve for \(x\):

For \(u = 1\), \(\cos \frac{x}{2} = 1\), so \(\frac{x}{2} = 0\) or \(x = 0\).

For \(u = -\frac{3}{4}\), \(\cos \frac{x}{2} = -\frac{3}{4}\), solve for \(\frac{x}{2}\):

\(\frac{x}{2} = \cos^{-1}(-\frac{3}{4})\), so \(x = 2 \cos^{-1}(-\frac{3}{4})\).

9. Calculate \(x\) for \(\cos^{-1}(-\frac{3}{4})\):

\(x \approx 4.84\).

Thus, the solutions are \(x = 0\) and \(x \approx 4.84\).

To solve the equation \(\csc 2\theta = \sec \theta + \cot \theta\), we start by expressing all terms in terms of sine and cosine:

\(\csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2\sin \theta \cos \theta}\)

\(\sec \theta = \frac{1}{\cos \theta}\)

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

Substitute these into the equation:

\(\frac{1}{2\sin \theta \cos \theta} = \frac{1}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\)

Multiply through by \(2\sin \theta \cos \theta\) to clear the fractions:

\(1 = 2\sin \theta + 2\cos^2 \theta\)

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

\(1 = 2\sin \theta + 2(1 - \sin^2 \theta)\)

Simplify:

\(1 = 2\sin \theta + 2 - 2\sin^2 \theta\)

\(2\sin^2 \theta - 2\sin \theta - 1 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

\(2x^2 - 2x - 1 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = -2, c = -1\):

\(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{4}\)

\(x = \frac{2 \pm \sqrt{4 + 8}}{4}\)

\(x = \frac{2 \pm \sqrt{12}}{4}\)

\(x = \frac{2 \pm 2\sqrt{3}}{4}\)

\(x = \frac{1 \pm \sqrt{3}}{2}\)

Thus, \(\sin \theta = \frac{1 + \sqrt{3}}{2}\) or \(\sin \theta = \frac{1 - \sqrt{3}}{2}\).

For \(\sin \theta = \frac{1 + \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 201.5^\circ\).

For \(\sin \theta = \frac{1 - \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 338.5^\circ\).

Thus, the solutions are \(\theta = 201.5^\circ\) and \(\theta = 338.5^\circ\).

To solve the equation \(\cos \theta + 4 \cos 2\theta = 3\), we start by using the double angle identity \(\cos 2\theta = 2\cos^2\theta - 1\).

Substitute \(\cos 2\theta\) in the equation:

\(\cos \theta + 4(2\cos^2\theta - 1) = 3\)

Simplify the equation:

\(\cos \theta + 8\cos^2\theta - 4 = 3\)

Rearrange terms:

\(8\cos^2\theta + \cos \theta - 7 = 0\)

This is a quadratic equation in \(\cos \theta\). Let \(x = \cos \theta\), then:

\(8x^2 + x - 7 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = 1\), \(c = -7\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 8 \times (-7)}}{2 \times 8}\)

\(x = \frac{-1 \pm \sqrt{1 + 224}}{16}\)

\(x = \frac{-1 \pm \sqrt{225}}{16}\)

\(x = \frac{-1 \pm 15}{16}\)

So, \(x = \frac{14}{16} = \frac{7}{8}\) or \(x = \frac{-16}{16} = -1\).

Thus, \(\cos \theta = \frac{7}{8}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{7}{8}\):

\(\theta = \cos^{-1}\left(\frac{7}{8}\right) \approx 29^\circ\)

For \(\cos \theta = -1\):

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 29^\circ\) and \(\theta = 180^\circ\).

1. Start with the equation \(\sin \theta = 2 \cos 2\theta + 1\).

2. Use the double angle formula for cosine: \(\cos 2\theta = 1 - 2\sin^2 \theta\).

3. Substitute into the equation: \(\sin \theta = 2(1 - 2\sin^2 \theta) + 1\).

4. Simplify: \(\sin \theta = 2 - 4\sin^2 \theta + 1\).

5. Rearrange to form a quadratic equation: \(4\sin^2 \theta + \sin \theta - 3 = 0\).

6. Solve the quadratic equation using the quadratic formula: \(\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = 1, c = -3\).

7. Calculate the discriminant: \(b^2 - 4ac = 1 + 48 = 49\).

8. Solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm 7}{8}\)

9. This gives \(\sin \theta = \frac{3}{4}\) or \(\sin \theta = -1\).

10. For \(\sin \theta = \frac{3}{4}\), find \(\theta\):

\(\theta = \sin^{-1}\left(\frac{3}{4}\right) \approx 48.6^\circ\) and \(180^\circ - 48.6^\circ = 131.4^\circ\).

11. For \(\sin \theta = -1\), \(\theta = 270^\circ\).

12. The solutions in the interval \(0^\circ \leq \theta \leq 360^\circ\) are \(\theta = 48.6^\circ, 131.4^\circ, 270^\circ\).