Solve the following equation
`(i) 5cos2theta+2cos^(2)"(theta)/(2)+1=0`, `-(pi)/(2) lt theta lt (pi)/(2)`
`(ii) sin7theta+sin4theta+sintheta=0`, `0 le theta le pi`
`(iii) tantheta+sectheta=sqrt(3)`, `0 le theta le 2pi`

Answer

Let's solve the given equations step by step.

### (i) Solve the equation:
\[ 5 \cos(2\theta) + 2 \cos^2\left(\frac{\theta}{2}\right) + 1 = 0 \]

**Step 1:** Use the identity for \(\cos(2\theta)\):
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Substituting this into the equation gives:
\[ 5(2\cos^2(\theta) - 1) + 2\cos^2\left(\frac{\theta}{2}\right) + 1 = 0 \]

**Step 2:** Simplify the equation:
\[ 10\cos^2(\theta) - 5 + 2\cos^2\left(\frac{\theta}{2}\right) + 1 = 0 \]
\[ 10\cos^2(\theta) + 2\cos^2\left(\frac{\theta}{2}\right) - 4 = 0 \]

**Step 3:** Substitute \(\cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos(\theta)}{2}\):
\[ 10\cos^2(\theta) + 2\left(\frac{1 + \cos(\theta)}{2}\right) - 4 = 0 \]
\[ 10\cos^2(\theta) + 1 + \cos(\theta) - 4 = 0 \]
\[ 10\cos^2(\theta) + \cos(\theta) - 3 = 0 \]

**Step 4:** Let \(x = \cos(\theta)\):
\[ 10x^2 + x - 3 = 0 \]

**Step 5:** Solve the quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 120}}{20} = \frac{-1 \pm 11}{20} \]
Calculating gives:
\[ x_1 = \frac{10}{20} = \frac{1}{2}, \quad x_2 = \frac{-12}{20} = -\frac{3}{5} \]

**Step 6:** Find \(\theta\):
1. For \(x_1 = \frac{1}{2}\):
\[ \cos(\theta) = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \]
2. For \(x_2 = -\frac{3}{5}\):
\[ \cos(\theta) = -\frac{3}{5} \Rightarrow \theta \text{ is not in } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]

**Final Answer for (i):**
\[ \theta = \frac{\pi}{3} \]

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### (ii) Solve the equation:
\[ \sin(7\theta) + \sin(4\theta) + \sin(\theta) = 0 \]

**Step 1:** Group \(\sin(7\theta)\) and \(\sin(\theta)\):
Using the identity \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\):
\[ \sin(7\theta) + \sin(\theta) = 2 \sin(4\theta) \cos(3\theta) \]
Thus, the equation becomes:
\[ 2 \sin(4\theta) \cos(3\theta) + \sin(4\theta) = 0 \]

**Step 2:** Factor out \(\sin(4\theta)\):
\[ \sin(4\theta)(2\cos(3\theta) + 1) = 0 \]

**Step 3:** Solve each factor:
1. \(\sin(4\theta) = 0\):
\[ 4\theta = n\pi \Rightarrow \theta = \frac{n\pi}{4}, \quad n = 0, 1, 2, 3, 4 \]
Valid solutions in \([0, \pi]\) are:
\(\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\)

2. \(2\cos(3\theta) + 1 = 0\):
\[ \cos(3\theta) = -\frac{1}{2} \Rightarrow 3\theta = \frac{2\pi}{3}, \frac{4\pi}{3} \Rightarrow \theta = \frac{2\pi}{9}, \frac{4\pi}{9} \]

**Final Answer for (ii):**
\[ \theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{2\pi}{9}, \frac{4\pi}{9} \]

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### (iii) Solve the equation:
\[ \tan(\theta) + \sec(\theta) = \sqrt{3} \]

**Step 1:** Rewrite \(\tan(\theta)\) and \(\sec(\theta)\):
\[ \frac{\sin(\theta)}{\cos(\theta)} + \frac{1}{\cos(\theta)} = \sqrt{3} \]
Multiply through by \(\cos(\theta)\):
\[ \sin(\theta) + 1 = \sqrt{3}\cos(\theta) \]

**Step 2:** Rearrange:
\[ \sin(\theta) - \sqrt{3}\cos(\theta) + 1 = 0 \]

**Step 3:** Square both sides:
\[ (\sin(\theta) - \sqrt{3}\cos(\theta))^2 = 1 \]
Expanding gives:
\[ \sin^2(\theta) - 2\sqrt{3}\sin(\theta)\cos(\theta) + 3\cos^2(\theta) = 1 \]

**Step 4:** Use \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\[ 1 - 2\sqrt{3}\sin(\theta)\cos(\theta) + 3\cos^2(\theta) = 1 \]
This simplifies to:
\[ -2\sqrt{3}\sin(\theta)\cos(\theta) + 3\cos^2(\theta) = 0 \]

**Step 5:** Factor out \(\cos(\theta)\):
\[ \cos(\theta)(3\cos(\theta) - 2\sqrt{3}\sin(\theta)) = 0 \]

**Step 6:** Solve:
1. \(\cos(\theta) = 0\):
\[ \theta = \frac{\pi}{2}, \frac{3\pi}{2} \]
2. \(3\cos(\theta) = 2\sqrt{3}\sin(\theta)\):
\[ \tan(\theta) = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6}, \frac{7\pi}{6} \]

**Final Answer for (iii):**
\[ \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{7\pi}{6} \]

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