To solve the trigonometric equation \( \cos(2x) - 7\sin x - 4 = 0 \), we can use trigonometric identities to express everything in terms of \(\sin x\) and solve for \(x\).

### Step-by-Step Solution

1. **Use the Double Angle Identity:**

   The double angle identity for cosine is:
   \[
   \cos(2x) = 1 - 2\sin^2 x
   \]
   Substituting this into the equation gives:
   \[
   1 - 2\sin^2 x - 7\sin x - 4 = 0
   \]

2. **Simplify the Equation:**

   \[
   -2\sin^2 x - 7\sin x + 1 - 4 = 0
   \]
   \[
   -2\sin^2 x - 7\sin x - 3 = 0
   \]

3. **Multiply Through by \(-1\):**

   To make the equation easier to handle, multiply everything by \(-1\):
   \[
   2\sin^2 x + 7\sin x + 3 = 0
   \]

4. **Let \( u = \sin x \):**

   Substitute \( u \) for \( \sin x \):
   \[
   2u^2 + 7u + 3 = 0
   \]

5. **Solve the Quadratic Equation:**

   This is a quadratic equation in standard form \( au^2 + bu + c = 0 \), where \( a = 2 \), \( b = 7 \), and \( c = 3 \).

   To find the roots, use the quadratic formula:
   \[
   u = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
   \]
   Substitute the values:
   \[
   u = \frac{{-7 \pm \sqrt{{7^2 - 4 \cdot 2 \cdot 3}}}}{2 \cdot 2}
   \]
   \[
   u = \frac{{-7 \pm \sqrt{{49 - 24}}}}{4}
   \]
   \[
   u = \frac{{-7 \pm \sqrt{25}}}{4}
   \]
   \[
   u = \frac{{-7 \pm 5}}{4}
   \]

   Calculate the two possible values for \( u \):
   \[
   u_1 = \frac{{-7 + 5}}{4} = \frac{{-2}}{4} = -\frac{1}{2}
   \]
   \[
   u_2 = \frac{{-7 - 5}}{4} = \frac{{-12}}{4} = -3
   \]

   Since \( \sin x \) must be between -1 and 1, the solution \(\sin x = -3\) is not valid. We are left with:
   \[
   \sin x = -\frac{1}{2}
   \]

6. **Find \( x \):**

   Now, solve for \( x \) when \(\sin x = -\frac{1}{2}\).

   The general solution for \(\sin x = -\frac{1}{2}\) is:
   \[
   x = \arcsin\left(-\frac{1}{2}\right) + 2k\pi
   \]
   \[
   x = -\frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = -\frac{5\pi}{6} + 2k\pi
   \]

   Since \(\sin x = -\frac{1}{2}\), the angles corresponding in the unit circle are:
   \[
   x = -\frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = -\frac{5\pi}{6} + 2k\pi
   \]

   Therefore, the solutions are:
   \[
   x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = -\frac{5\pi}{6} + 2k\pi
   \]

7. **Verify the Solutions:**

   Let's confirm the solutions satisfy the original equation:
   
   - For \(x = -\frac{\pi}{6} + 2k\pi\):
     \[
     \cos\left(2\left(-\frac{\pi}{6} + 2k\pi\right)\right) = \cos\left(-\frac{\pi}{3} + 4k\pi\right) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}
     \]
     \[
     7\sin\left(-\frac{\pi}{6} + 2k\pi\right) = 7\left(-\frac{1}{2}\right) = -\frac{7}{2}
     \]
     \[
     \cos(2x) - 7\sin x - 4 = \frac{1}{2} - \frac{7}{2} - 4 = 0
     \]
     Thus, \(x = -\frac{\pi}{6} + 2k\pi\) satisfies the equation.

   - For \(x = -\frac{5\pi}{6} + 2k\pi\):
     \[
     \cos\left(2\left(-\frac{5\pi}{6} + 2k\pi\right)\right) = \cos\left(-\frac{5\pi}{3} + 4k\pi\right) = \cos\left(-\frac{5\pi}{3}\right) = \frac{1}{2}
     \]
     \[
     7\sin\left(-\frac{5\pi}{6} + 2k\pi\right) = 7\left(-\frac{1}{2}\right) = -\frac{7}{2}
     \]
     \[
     \cos(2x) - 7\sin x - 4 = \frac{1}{2} - \frac{7}{2} - 4 = 0
     \]
     Thus, \(x = -\frac{5\pi}{6} + 2k\pi\) also satisfies the equation.

Therefore, the solutions are:
\[
x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = -\frac{5\pi}{6} + 2k\pi \quad \text{for any integer } k
\]

### Solution Summary
The equation \( \cos(2x) - 7\sin x - 4 = 0 \) has solutions:
\[
x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = -\frac{5\pi}{6} + 2k\pi
\]
where \(k\) is any integer.

### Additional Questions
1. What are the solutions for \(x\) if \(\sin x = \frac{1}{2}\)?
2. How do we derive the general solution for \(\cos(2x) = \sin x\)?
3. What are the general solutions for \(\tan x = 1\)?
4. How do you solve \( \cos^2 x - \sin x - 1 = 0 \)?
5. What is the effect of adding a constant to a trigonometric equation?
6. How can we solve equations involving \(\cos x + \sin x\)?
7. What are the solutions for \(\sin 2x = 0.5\)?
8. How do we graphically represent \(\cos(2x) - 7\sin x - 4 = 0\)?

### Tip
When solving trigonometric equations, always remember to check for extraneous solutions that might arise when squaring or applying identities.

Learn how to solve the trigonometric equation cos(2x) - 7sinx - 4 = 0 with detailed step-by-step instructions. This problem involves using trigonometric identities and the quadratic formula, suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation