If $4\sin^2 x - 3 = 0$, find the value of x, when 0° $\leq$ x $\leq$ 90°

1. Analyze the question

We are given a trigonometric equation $4\sin^2 x - 3 = 0$ and asked to find the value of $x$ within the range $0^\circ \leq x \leq 90^\circ$.

2. Explain the concept

To solve this, we need to isolate the trigonometric function ($\sin x$) and then use the inverse sine function (or our knowledge of special angles) to find the value of $x$. Remember that $\sin^2 x$ is the same as $(\sin x)^2$.

3. Show the solution

Step 1: Isolate $\sin^2 x$ Start with the equation: $4\sin^2 x - 3 = 0$

Add 3 to both sides: $4\sin^2 x = 3$

Divide both sides by 4: $\sin^2 x = \frac{3}{4}$

Step 2: Solve for $\sin x$ Take the square root of both sides: $\sqrt{\sin^2 x} = \sqrt{\frac{3}{4}}$ $\sin x = \frac{\sqrt{3}}{\sqrt{4}}$ $\sin x = \frac{\sqrt{3}}{2}$

Step 3: Find the value of $x$ We need to find the angle $x$ whose sine is $\frac{\sqrt{3}}{2}$. From the standard trigonometric ratios for special angles: $\sin 60^\circ = \frac{\sqrt{3}}{2}$

Since $60^\circ$ falls within the given range ($0^\circ \leq x \leq 90^\circ$), this is our solution.

4. Identify the answer

Comparing our result to the options provided: A. 90° B. 45° C. 60° D. 30°

The correct option is C.

5. Tip

Always memorize the "Special Angles" table for $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$. A quick way to remember the sine values is:

• $\sin 0^\circ = \sqrt{0}/2 = 0$
• $\sin 30^\circ = \sqrt{1}/2 = 1/2$
• $\sin 45^\circ = \sqrt{2}/2$
• $\sin 60^\circ = \sqrt{3}/2$
• $\sin 90^\circ = \sqrt{4}/2 = 1$