The number of values of `theta` in `[0, 2pi]` that satisfy the equation `3cos2theta+13sin theta-8=0` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \(3\cos(2\theta) + 13\sin(\theta) - 8 = 0\) for the number of values of \(\theta\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Rewrite \(\cos(2\theta)\) in terms of \(\sin(\theta)\)
We know that:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Substituting this into the equation gives:
\[ 3(1 - 2\sin^2(\theta)) + 13\sin(\theta) - 8 = 0 \]

### Step 2: Simplify the equation
Expanding the equation:
\[ 3 - 6\sin^2(\theta) + 13\sin(\theta) - 8 = 0 \]
This simplifies to:
\[ -6\sin^2(\theta) + 13\sin(\theta) - 5 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ 6\sin^2(\theta) - 13\sin(\theta) + 5 = 0 \]

### Step 3: Let \(t = \sin(\theta)\)
Substituting \(t\) for \(\sin(\theta)\), we have:
\[ 6t^2 - 13t + 5 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 6\), \(b = -13\), and \(c = 5\):
\[ t = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 6 \cdot 5}}{2 \cdot 6} \]
Calculating the discriminant:
\[ t = \frac{13 \pm \sqrt{169 - 120}}{12} \]
\[ t = \frac{13 \pm \sqrt{49}}{12} \]
\[ t = \frac{13 \pm 7}{12} \]
This gives us two potential solutions:
\[ t_1 = \frac{20}{12} = \frac{5}{3} \quad \text{(not valid, as } \sin(\theta) \text{ cannot exceed 1)} \]
\[ t_2 = \frac{6}{12} = \frac{1}{2} \]

### Step 5: Find \(\theta\) values for valid \(t\)
Since \(t = \sin(\theta)\), we have:
\[ \sin(\theta) = \frac{1}{2} \]
The solutions for \(\theta\) in the interval \([0, 2\pi]\) are:
\[ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} \]

### Step 6: Count the number of solutions
Thus, there are **2 values of \(\theta\)** that satisfy the equation in the interval \([0, 2\pi]\).

### Final Answer
The number of values of \(\theta\) in \([0, 2\pi]\) that satisfy the equation is **2**.