Precalculus Trigonometric Equations and Identities Study Guide

Q1. Solve for all values of $\theta$ in the equation: $2 - \sqrt{3} \csc(\theta) = 0$. Give a general formula for all solutions.

Background

Topic: Solving Trigonometric Equations

This question tests your ability to solve basic trigonometric equations and express all possible solutions using general solution formulas.

Key Terms and Formulas:

• $\csc(\theta)$ is the cosecant function, which is the reciprocal of $\sin(\theta)$: $\csc(\theta) = \frac{1}{\sin(\theta)}$

• General solutions for trigonometric equations often involve adding integer multiples of the period (e.g., $2\pi k$ for sine/cosine, where $k$ is any integer).

Step-by-Step Guidance

1. Start by isolating $\csc(\theta)$ in the equation: $2 - \sqrt{3} \csc(\theta) = 0$.

2. Rearrange to solve for $\csc(\theta)$: $\sqrt{3} \csc(\theta) = 2$.

3. Divide both sides by $\sqrt{3}$ to get $\csc(\theta) = \frac{2}{\sqrt{3}}$.

4. Recall that $\csc(\theta) = \frac{1}{\sin(\theta)}$, so set $\frac{1}{\sin(\theta)} = \frac{2}{\sqrt{3}}$ and solve for $\sin(\theta)$.

Try solving on your own before revealing the answer!

Q2. Solve for all values of $\theta$ in the equation: $\sin\left(\frac{\theta}{2}\right) = -\frac{\sqrt{3}}{2}$.

Background

Topic: Solving Trigonometric Equations with Half-Angle Arguments

This question tests your ability to solve equations involving sine with a half-angle argument and to find all solutions.

Key Terms and Formulas:

• Recall the unit circle values for $\sin(x) = -\frac{\sqrt{3}}{2}$.

• General solutions for $\sin(x) = a$ are $x = \arcsin(a) + 2\pi k$ and $x = \pi - \arcsin(a) + 2\pi k$.

Step-by-Step Guidance

1. Let $x = \frac{\theta}{2}$, so the equation becomes $\sin(x) = -\frac{\sqrt{3}}{2}$.

2. Determine all angles $x$ where $\sin(x) = -\frac{\sqrt{3}}{2}$ (think about which quadrants sine is negative and the reference angle).

3. Write the general solutions for $x$ using the arcsin and the period of sine.

4. Substitute back $x = \frac{\theta}{2}$ and solve for $\theta$.

Try solving on your own before revealing the answer!

Q3. Solve the equation $2\sin(\theta) + 3 = 2$ for $0 \leq \theta < 2\pi$.

Background

Topic: Solving Basic Trigonometric Equations on a Restricted Interval

This question asks you to solve a linear trigonometric equation for all solutions within one period ($0 \leq \theta < 2\pi$).

Key Terms and Formulas:

• Isolate the trigonometric function first.

• Recall the solutions for $\sin(\theta) = a$ on $[0, 2\pi)$.

Step-by-Step Guidance

1. Subtract 3 from both sides: $2\sin(\theta) = 2 - 3$.

2. Simplify to $2\sin(\theta) = -1$.

3. Divide both sides by 2: $\sin(\theta) = -\frac{1}{2}$.

4. Find all $\theta$ in $[0, 2\pi)$ where $\sin(\theta) = -\frac{1}{2}$ (think about the unit circle and which quadrants sine is negative).

Try solving on your own before revealing the answer!

Q4. Solve the equation $3\sqrt{2}\cos(\theta) + 2 = -1$ for $0 \leq \theta < 2\pi$.

Background

Topic: Solving Linear Trigonometric Equations

This question tests your ability to solve for $\theta$ when the equation involves cosine and a square root coefficient.

Key Terms and Formulas:

• Isolate $\cos(\theta)$ first.

• Recall the solutions for $\cos(\theta) = a$ on $[0, 2\pi)$.

Step-by-Step Guidance

1. Subtract 2 from both sides: $3\sqrt{2}\cos(\theta) = -1 - 2$.

2. Simplify to $3\sqrt{2}\cos(\theta) = -3$.

3. Divide both sides by $3\sqrt{2}$: $\cos(\theta) = -\frac{1}{\sqrt{2}}$.

4. Find all $\theta$ in $[0, 2\pi)$ where $\cos(\theta) = -\frac{1}{\sqrt{2}}$ (think about the unit circle and which quadrants cosine is negative).

Try solving on your own before revealing the answer!

Q5. Solve the quadratic equation $2\sin^2(\theta) - \sin(\theta) - 1 = 0$ for $0 \leq \theta < 2\pi$.

Background

Topic: Solving Quadratic Trigonometric Equations

This question tests your ability to factor or use the quadratic formula to solve trigonometric equations.

Key Terms and Formulas:

• Let $x = \sin(\theta)$ to rewrite the equation as a quadratic in $x$.

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step-by-Step Guidance

1. Let $x = \sin(\theta)$ and rewrite the equation: $2x^2 - x - 1 = 0$.

2. Factor or use the quadratic formula to solve for $x$.

3. Once you have the possible values for $\sin(\theta)$, determine which values are valid for $\theta$ in $[0, 2\pi)$.

4. Find all $\theta$ corresponding to those sine values within the given interval.

Try solving on your own before revealing the answer!