solve using trignometric identities: 2cos^2x+5sinx=4

To solve the trigonometric equation \( 2\cos^2x + 5\sin x = 4 \) using trigonometric identities, we proceed as follows:

### Step 1: Express \( \cos^2x \) in terms of \( \sin x \)
Use the Pythagorean identity:
\[
\cos^2x = 1 - \sin^2x
\]
Substitute this into the original equation:
\[
2(1 - \sin^2x) + 5\sin x = 4
\]
This simplifies to:
\[
2 - 2\sin^2x + 5\sin x = 4
\]

### Step 2: Rearrange the equation
Move all terms to one side:
\[
-2\sin^2x + 5\sin x + 2 - 4 = 0
\]
\[
-2\sin^2x + 5\sin x - 2 = 0
\]
Multiply the entire equation by \(-1\) to make the quadratic coefficient positive:
\[
2\sin^2x - 5\sin x + 2 = 0
\]

### Step 3: Solve the quadratic equation
Let \( u = \sin x \). The equation becomes:
\[
2u^2 - 5u + 2 = 0
\]
This is a quadratic equation in \( u \). Use the quadratic formula to solve for \( u \):
\[
u = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}
\]
\[
u = \frac{5 \pm \sqrt{25 - 16}}{4}
\]
\[
u = \frac{5 \pm \sqrt{9}}{4}
\]
\[
u = \frac{5 \pm 3}{4}
\]
This gives us two possible values for \( u \):
\[
u = \frac{8}{4} = 2 \quad (\text{not valid since } \sin x \text{ must be between } -1 \text{ and } 1)
\]
\[
u = \frac{2}{4} = \frac{1}{2}
\]

### Step 4: Find the values of \( x \)
Now, solve for \( x \) when \( \sin x = \frac{1}{2} \).

The general solution for \( \sin x = \frac{1}{2} \) is:
\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi
\]
where \( n \) is any integer.

### Final Answer:
The solutions are:
\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi \quad \text{for any integer } n.
\]

Would you like further clarification, or do you have any questions?

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**Here are 5 related questions:**

1. How can you use the Pythagorean identity in trigonometric equations?
2. What is the significance of the unit circle in solving trigonometric equations?
3. How do you derive the general solution for trigonometric equations involving \( \sin x \) or \( \cos x \)?
4. What are some common mistakes to avoid when solving trigonometric equations?
5. How do you solve trigonometric equations that involve both sine and cosine functions?

**Tip:** Always double-check the range of your trigonometric functions to ensure your solutions are valid within the given domain.

Learn how to solve the trigonometric equation 2cos^2x + 5sinx = 4 step-by-step. This problem involves using trigonometric identities and the quadratic formula to find solutions. Suitable for students in Grades 10-12.

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