Solve `2cos^2x+3sinx=0`.

Options

A. nπ+π/6
B. nπ-π/6
C. nπ+((-1)^n)π/6
D. nπ+((-1)^(n+1))π/6

Answer

To solve the equation \( 2\cos^2 x + 3\sin x = 0 \), we can follow these steps:

### Step 1: Rewrite \(\cos^2 x\) in terms of \(\sin x\)
Using the Pythagorean identity, we can express \(\cos^2 x\) as:
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into the equation gives:
\[ 2(1 - \sin^2 x) + 3\sin x = 0 \]

### Step 2: Simplify the equation
Distributing the \(2\) in the equation:
\[ 2 - 2\sin^2 x + 3\sin x = 0 \]
Rearranging this, we get:
\[ -2\sin^2 x + 3\sin x + 2 = 0 \]

### Step 3: Multiply through by -1
To make the leading coefficient positive, we multiply the entire equation by \(-1\):
\[ 2\sin^2 x - 3\sin x - 2 = 0 \]

### Step 4: Factor the quadratic equation
We need to factor the quadratic equation \(2\sin^2 x - 3\sin x - 2 = 0\). We can rewrite \(-3\sin x\) as \(-4\sin x + \sin x\):
\[ 2\sin^2 x - 4\sin x + \sin x - 2 = 0 \]
Grouping the terms gives:
\[ (2\sin^2 x - 4\sin x) + (\sin x - 2) = 0 \]
Factoring out common terms:
\[ 2\sin x(\sin x - 2) + 1(\sin x - 2) = 0 \]
This can be factored as:
\[ (2\sin x + 1)(\sin x - 2) = 0 \]

### Step 5: Solve for \(\sin x\)
Setting each factor to zero gives us two equations:
1. \(2\sin x + 1 = 0\)
2. \(\sin x - 2 = 0\)

For the first equation:
\[ 2\sin x + 1 = 0 \implies \sin x = -\frac{1}{2} \]

For the second equation:
\[ \sin x - 2 = 0 \implies \sin x = 2 \]
However, \(\sin x = 2\) is not possible since the range of \(\sin x\) is \([-1, 1]\).

### Step 6: Find the general solution for \(\sin x = -\frac{1}{2}\)
The solutions for \(\sin x = -\frac{1}{2}\) occur at:
\[ x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi \]
where \(n\) is any integer.

### Final Solution
Thus, the general solution for the equation \(2\cos^2 x + 3\sin x = 0\) is:
\[ x = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2n\pi, \quad n \in \mathbb{Z} \]

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