The equation presented is:

$6\sin^2 x + 5\cos x - 2 = 0.$

Let's solve this step-by-step.

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Step 1: Express everything in terms of $\cos x$ and $\sin^2 x$

Using the trigonometric identity:

$\sin^2 x = 1 - \cos^2 x,$

we substitute into the equation:

$6(1 - \cos^2 x) + 5\cos x - 2 = 0.$

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Step 2: Simplify the expression

Distribute the $6$ over the terms inside the parenthesis:

$6 - 6\cos^2 x + 5\cos x - 2 = 0.$

Combine like terms:

$-6\cos^2 x + 5\cos x + 4 = 0.$

Multiply through by $-1$ to make the quadratic simpler:

$6\cos^2 x - 5\cos x - 4 = 0.$

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Step 3: Solve the quadratic equation

Let $y = \cos x$. The equation becomes:

$6y^2 - 5y - 4 = 0.$

We solve this quadratic using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$

where $a = 6$, $b = -5$, and $c = -4$:

$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-4)}}{2 \cdot 6}.$

Simplify:

$y = \frac{5 \pm \sqrt{25 + 96}}{12}.$

$y = \frac{5 \pm \sqrt{121}}{12}.$

$y = \frac{5 \pm 11}{12}.$

Thus, we have two solutions:

1. $y = \frac{16}{12} = \frac{4}{3}$
2. $y = \frac{-6}{12} = -\frac{1}{2}$

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Step 4: Interpret the solutions

1. $\cos x = \frac{4}{3}$: This is not possible since $\cos x$ must be in the range $[-1, 1]$. Therefore, discard this solution.

2. $\cos x = -\frac{1}{2}$: This is valid.

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Step 5: Solve for $x$

The general solution for $\cos x = -\frac{1}{2}$ is:

$x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi,$

where $k$ is any integer.

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Final Answer

The solutions to the equation are:

$x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}.$

Would you like further clarification or a detailed explanation on any step?

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Related Questions

1. How do you solve trigonometric equations involving quadratic forms?
2. What is the quadratic formula and how does it apply to trigonometric functions?
3. How can you convert between $\sin^2 x$ and $\cos^2 x$ using identities?
4. What are the restrictions on the values of $\cos x$ and $\sin x$?
5. How do you determine the general solutions of trigonometric equations?

Tip: When solving trigonometric equations, always check if the solutions fall within the valid range of the trigonometric functions.