Question

Solve \(4 \sin ^{2} x=1\) algebraically over the domain \(-180^{\circ} \leq x<180^{\circ}\).

Short answer

The solutions are \(x = 30^{\circ}, -30^{\circ}, 150^{\circ}, \text{and} -150^{\circ} \).

Step-by-step solution

Isolate \( \sin^{2} x \) term

Given the equation \(4 \sin^{2} x = 1\), divide both sides by 4 to isolate \(\sin^{2} x \): \[ \sin^{2} x = \frac{1}{4} \]

Take the square root

Take the square root of both sides to solve for \(\sin x\): \[ \sin x = \pm \frac{1}{2} \]

Find the angle corresponding to \( \sin x = \frac{1}{2} \)

Determine the angles in the given domain where \( \sin x = \frac{1}{2} \): \[ x = 30^{\circ}, 150^{\circ} \] Since the domain is \( -180^{\circ} \leq x < 180^{\circ} \), we also need the corresponding negative angles: \[ x = -30^{\circ}, -150^{\circ} \]

Find the angle corresponding to \( \sin x = \frac{-1}{2} \)

Determine the angles in the given domain where \( \sin x = \frac{-1}{2} \): \[ x = -30^{\circ}, -150^{\circ} \] Since the domain is \( -180^{\circ} \leq x < 180^{\circ} \), we also need the corresponding positive angles: \[ x = 210^{\circ}, -210^{\circ} \] Simplify to fit within the domain: \[ x = 30^{\circ}, -30^{\circ}, 150^{\circ}, -150^{\circ} \]

Key concepts

solving trigonometric equations

When it comes to solving trigonometric equations, you'll typically follow a set of systematic steps to find the solutions. Trigonometric equations often involve \( \sin, \cos, \tan \) and their inverses. Being methodical is key.
First, you need to isolate the trigonometric function. This often involves algebraic manipulation, similar to solving any other algebraic equation.
Once isolated, identify the general solutions to the trigonometric function. This might require knowledge of key angles or using tools like the unit circle.

After finding the general solutions, you'll need to identify specific solutions within a given domain. This usually involves understanding the periodic nature of trigonometric functions.

• Eliminate extraneous solutions by considering the specifics of the domain.

Hence, mastery in solving trigonometric equations comes from practice and familiarity with the fundamental properties and transformations of trigonometric functions.

isolating trigonometric functions

Isolating trigonometric functions is an essential step when solving trigonometric equations. This involves using algebraic techniques to get the trigonometric term by itself.
In our given example, the equation is \(4 \sin^{2} x = 1\). We divide both sides by 4, thus isolating \(\sin^{2} x\):

\[ \sin^{2} x = \frac{1}{4} \]

This transformation simplifies the equation and prepares it for further manipulation.

• Understand that we'll need to take the square root of both sides to solve for \( \sin x \).
• Remember, taking the square root introduces both positive and negative roots. Hence, \( \sin x = \frac{1}{2}\) and \( \sin x = \frac{-1}{2}\).

Mastering the skill of isolating trigonometric functions can make solving these equations much more straightforward.

unit circle

The unit circle is a crucial tool in trigonometry, especially when solving trigonometric equations. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate system.

Each point on the circle corresponds to an angle whose sine and cosine values can be read directly from the coordinates of the point. For instance, at \(30^{\circ} \), the coordinates are \(( \frac{ \sqrt{3}}{2}, \frac{1}{2} \))\. Therefore, \(\sin(30^{\circ}) = \frac{1}{2} \).
Similarly:

• \( \sin(150^{\circ}) = \frac{1}{2} \)
• \( \sin(-30^{\circ}) = -\frac{1}{2} \)
• \( \sin(-150^{\circ}) = -\frac{1}{2} \)

Using the unit circle helps identify these key angles quickly and accurately, facilitating the solving process.

positive and negative angles

When solving trigonometric equations, recognizing positive and negative angles is vital.

• Positive angles are measured counterclockwise from the positive x-axis.
• Negative angles are measured clockwise from the positive x-axis.

Consider the equation \(\sin x = \frac{1}{2}\):
\( x = 30^{\circ} \) and \( x = 150^{\circ} \) are solutions for positive angles within the domain.
For negative angles, the corresponding solutions are:

• \( x = -30^{\circ} \) and \( x = -150^{\circ} \).

Handling both positive and negative angles ensures you find all possible solutions within the specified domain. This complete understanding is crucial for tackling any trigonometric equation efficiently.