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The problem in the image is to solve for \( x \) in the equation:

\[
16 \sin^2 x = 1
\]

for the range \( 0^\circ \leq x \leq 270^\circ \).

Let's break down the solution.

### Step 1: Simplify the Equation
Divide both sides by 16:

\[
\sin^2 x = \frac{1}{16}
\]

Take the square root of both sides, keeping in mind that \( \sin x \) can be either positive or negative:

\[
\sin x = \pm \frac{1}{4}
\]

### Step 2: Solve for \( x \)
Now, we need to find the angles \( x \) within the range \( 0^\circ \leq x \leq 270^\circ \) for which \( \sin x = \frac{1}{4} \) and \( \sin x = -\frac{1}{4} \).

#### Case 1: \( \sin x = \frac{1}{4} \)
Using inverse sine, we get:

\[
x = \sin^{-1}\left(\frac{1}{4}\right) \approx 14.48^\circ
\]

Since the sine function is positive in both the first and second quadrants, another solution in the range is:

\[
x = 180^\circ - 14.48^\circ = 165.52^\circ
\]

#### Case 2: \( \sin x = -\frac{1}{4} \)
Similarly, we find:

\[
x = \sin^{-1}\left(-\frac{1}{4}\right) \approx -14.48^\circ
\]

However, we need to find values within \( 0^\circ \leq x \leq 270^\circ \). In the third and fourth quadrants, where sine is negative, we get:

\[
x = 180^\circ + 14.48^\circ = 194.48^\circ
\]
\[
x = 360^\circ - 14.48^\circ = 345.52^\circ
\]

Since \( 345.52^\circ \) is outside the specified range, we discard it.

### Solution Summary
The values of \( x \) that satisfy the equation within the range \( 0^\circ \leq x \leq 270^\circ \) are:

\[
x \approx 14.48^\circ, \quad 165.52^\circ, \quad 194.48^\circ
\]

Would you like more details, or do you have any questions?

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Here are five related questions for further practice:
1. Solve \( 9 \cos^2 x = 1 \) for \( 0^\circ \leq x \leq 360^\circ \).
2. Solve \( 4 \sin x \cos x = 1 \) within \( 0^\circ \leq x \leq 180^\circ \).
3. Find all solutions of \( \tan^2 x = 3 \) in \( 0^\circ \leq x \leq 360^\circ \).
4. Solve \( \cos 2x = -0.5 \) for \( 0^\circ \leq x \leq 360^\circ \).
5. Solve \( 3 \sin^2 x + 2 \sin x - 1 = 0 \) in the interval \( 0^\circ \leq x \leq 180^\circ \).

**Tip:** Always check which quadrant corresponds to positive or negative values for trigonometric functions within a specified range.

Solve 16 sin^2 x = 1 for 0° ≤ x ≤ 270°.

Learn how to solve the trigonometric equation 16 sin^2 x = 1 within the range 0° ≤ x ≤ 270°. This solution involves simplifying trigonometric expressions and applying the inverse sine function to find all valid solutions. Suitable for students in Grades 10-12.

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