Solving Basic Trigonometric Equations

✖️ 1. General solution templates for sine, cosine, and tangent

🔄 General Solution Templates

For $\sin x = a$ , solutions are $x = \arcsin(a) + 2\pi k$ and $x = \pi - \arcsin(a) + 2\pi k$ where $k$ is any integer.
For $\cos x = a$ , solutions are $x = \arccos(a) + 2\pi k$ and $x = -\arccos(a) + 2\pi k$ .
For $\tan x = a$ , solutions are $x = \arctan(a) + \pi k$ (tangent repeats every $\pi$ , not $2\pi$ ).
The period determines the interval added: sine and cosine use $2\pi k$ , tangent uses $\pi k$ .
Always include the integer $k$ to capture all infinitely many solutions.

Example: $\sin x = 0.5$ gives $x = \frac{\pi}{6} + 2\pi k$ and $x = \frac{5\pi}{6} + 2\pi k$ .

💡 Sine and cosine repeat every full circle ( $2\pi$ ), tangent repeats every half circle ( $\pi$ ).

1. General solution templates for sine, cosine, and tangent

General Solution Templates for Sine, Cosine, and Tangent

A general solution to a basic trigonometric equation captures all angles that satisfy the equation by accounting for the periodic nature of trigonometric functions.

Intuition: Since $\sin x$ , $\cos x$ , and $\tan x$ repeat their values at regular intervals, any solution recurs infinitely many times, shifted by the function's period.

Core Rules:

For $\sin x = a$ where $|a| \leq 1$ : if $x = \alpha$ is one solution, then $x = \alpha + 2\pi k$ and $x = (\pi - \alpha) + 2\pi k$ for integer $k$ .
For $\cos x = a$ where $|a| \leq 1$ : if $x = \alpha$ is one solution, then $x = \pm \alpha + 2\pi k$ .
For $\tan x = a$ : if $x = \alpha$ is one solution, then $x = \alpha + \pi k$ (tangent has period $\pi$ ).

Consequence: These templates generate infinitely many solutions unless the domain is restricted.

Example: Solve $\sin x = \frac{1}{2}$ . One solution is $x = \frac{\pi}{6}$ , so the general solutions are $x = \frac{\pi}{6} + 2\pi k$ and $x = \frac{5\pi}{6} + 2\pi k$ , $k \in \mathbb{Z}$ .

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Level 2

ADV: GENERALIZE

Solve the equation $\tan x = 1$ . Which of the following represents the general solution for any integer $k$ ?

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✖️ 2. Solving equations within a restricted domain

📏 Restricted Domain Solutions

Start by finding the general solution with $k$ as an integer.
Substitute integer values of $k$ (like $k = 0, 1, -1, 2, -2$ ) until solutions fall outside the given interval.
Common domains are $[0, 2\pi)$ or $[0^\circ, 360^\circ)$ for one full cycle.
Discard any solution that lies outside the specified range.
Check both solution families from the general template.

Example: For $\cos x = -\frac{1}{2}$ on $[0, 2\pi)$ , general solutions are $x = \frac{2\pi}{3} + 2\pi k$ and $x = \frac{4\pi}{3} + 2\pi k$ . With $k=0$ , we get $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ .

💡 Plug in $k = 0, 1, -1$ and keep only what fits the interval.

2. Solving equations within a restricted domain

Solving Equations Within a Restricted Domain

Restricted domain solutions are the subset of general solutions that lie within a specified interval, commonly $[0, 2\pi)$ or $[0^\circ, 360^\circ)$ .

Intuition: Instead of listing infinitely many solutions, we identify only those angles within the given range by evaluating the general solution for appropriate integer values of $k$ .

Core Rules:

First, find the general solution using periodicity templates.
Substitute integer values of $k$ (typically $k = 0, \pm 1, \pm 2, \ldots$ ) until all solutions within the domain are found.
Verify that each candidate satisfies the domain boundaries (e.g., $0 \leq x < 2\pi$ ).
Discard any solution outside the interval.

Consequence: This approach yields a finite list of solutions, essential for practical applications with bounded contexts.

Example: Solve $\cos x = -\frac{\sqrt{2}}{2}$ on $[0, 2\pi)$ . General solutions are $x = \frac{3\pi}{4} + 2\pi k$ and $x = \frac{5\pi}{4} + 2\pi k$ . For $k=0$ , we get $x = \frac{3\pi}{4}, \frac{5\pi}{4}$ (both in $[0, 2\pi)$ ).

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Level 3

ADV: STRATEGY

Solve the equation $\sin x = 1/2$ on the restricted domain $[0, 2\pi)$ .

Which of the following represents the complete set of solutions?

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✖️ 3. Factoring trigonometric expressions and solving via the zero-product property

✂️ Factoring and Zero-Product Property

If a trig equation can be factored, set each factor equal to zero separately.
Use identities like $\sin^2 x + \cos^2 x = 1$ or $1 + \tan^2 x = \sec^2 x$ to simplify before factoring.
The zero-product property states: if $A \cdot B = 0$ , then $A = 0$ or $B = 0$ .
Solve each resulting simple equation independently.
Combine all solutions from each factor.

Example: $2\sin x \cos x = 0$ factors to $\sin x = 0$ or $\cos x = 0$ , giving $x = 0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}$ on $[0, 2\pi)$ .

💡 Factor first, then solve each piece like separate mini-equations.

3. Factoring trigonometric expressions and solving via the zero-product property

Factoring Trigonometric Expressions and Solving via the Zero-Product Property

Factoring rewrites a trigonometric equation as a product of simpler expressions set equal to zero, enabling the use of the zero-product property: if $AB = 0$ , then $A = 0$ or $B = 0$ .

Intuition: Just as with algebraic polynomials, factoring breaks a complex equation into simpler sub-equations, each solved independently.

Core Rules:

Rearrange the equation so one side equals zero.
Factor out common trigonometric terms (e.g., $\sin x$ , $\cos x$ ) or recognize patterns like difference of squares.
Set each factor equal to zero and solve separately.
Combine all solutions from each factor, checking for domain restrictions.

Consequence: This method efficiently handles equations involving products or sums of trigonometric functions.

Example: Solve $2\sin x \cos x - \cos x = 0$ . Factor: $\cos x(2\sin x - 1) = 0$ . Thus $\cos x = 0$ gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$ , and $2\sin x - 1 = 0$ gives $\sin x = \frac{1}{2}$ , so $x = \frac{\pi}{6}, \frac{5\pi}{6}$ on $[0, 2\pi)$ .

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Level 2

ADV: STRATEGY

Solve the equation: $\sin^2(x) - \sin(x) = 0$ .

What are all the solutions in the interval $[0, 2\pi)$ ?

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✖️ 4. Quadratic substitution methods and checking for extraneous solutions

🔁 Quadratic Substitution (u-substitution)

For equations like $2\sin^2 x - \sin x - 1 = 0$ , let $u = \sin x$ to get a standard quadratic $2u^2 - u - 1 = 0$ .
Solve the quadratic for $u$ using factoring, the quadratic formula, or completing the square.
Substitute back to get $\sin x = u$ and solve for $x$ .
Check validity: $\sin x$ and $\cos x$ must lie in $[-1, 1]$ ; discard any $u$ outside this range.
Always verify solutions in the original equation to catch extraneous roots.

Example: $\cos^2 x - 3\cos x + 2 = 0$ becomes $u^2 - 3u + 2 = 0$ , so $u = 1$ or $u = 2$ . Since $\cos x = 2$ is impossible, only $\cos x = 1$ gives $x = 0 + 2\pi k$ .

💡 After solving for $u$ , reject any value outside $[-1, 1]$ for sine or cosine.

4. Quadratic substitution methods and checking for extraneous solutions

Quadratic Substitution Methods and Checking for Extraneous Solutions

Quadratic substitution transforms a trigonometric equation into a standard quadratic form by letting $u$ equal a trigonometric function (e.g., $u = \sin x$ ), solving for $u$ , then back-substituting.

Intuition: Equations like $2\sin^2 x - 3\sin x + 1 = 0$ resemble quadratic polynomials; substitution simplifies solving.

Core Rules:

Substitute $u = \sin x$ , $u = \cos x$ , or $u = \tan x$ to obtain a quadratic in $u$ .
Solve the quadratic (factoring, quadratic formula, etc.).
Check validity: For $\sin x$ and $\cos x$ , ensure $|u| \leq 1$ ; discard any $u$ outside this range (extraneous solutions).
Back-substitute valid $u$ values to find $x$ .

Consequence: Substitution streamlines complex equations but requires vigilance to reject impossible trigonometric values.

Example: Solve $2\sin^2 x - \sin x - 1 = 0$ . Let $u = \sin x$ : $2u^2 - u - 1 = 0$ factors as $(2u + 1)(u - 1) = 0$ , giving $u = -\frac{1}{2}$ or $u = 1$ . Both are valid since $|u| \leq 1$ .

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Level 2

ADV: STRATEGY

Given the equation $3\cos^2 x + 5\cos x - 2 = 0$ , what is the correct quadratic equation after substituting $u = \cos x$ ?

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✖️ 5. Applications: Finding specific times when an AC voltage reaches a peak or threshold

⚡ AC Voltage Timing Problems

AC voltage is modeled as $V(t) = V_0 \sin(\omega t + \phi)$ where $V_0$ is peak voltage, $\omega$ is angular frequency, and $\phi$ is phase shift.
To find when voltage reaches a threshold $V_1$ , solve $V_0 \sin(\omega t + \phi) = V_1$ .
Isolate the sine: $\sin(\omega t + \phi) = \frac{V_1}{V_0}$ , then use general solution templates.
Divide by $\omega$ and subtract $\phi$ to solve for time $t$ .
Restrict $t$ to physically meaningful intervals (e.g., $t \geq 0$ or one cycle).

Example: If $V(t) = 120\sin(100\pi t)$ and we want $V = 60$ volts, then $\sin(100\pi t) = 0.5$ , so $100\pi t = \frac{\pi}{6} + 2\pi k$ or $100\pi t = \frac{5\pi}{6} + 2\pi k$ , giving $t = \frac{1}{600} + \frac{k}{50}$ or $t = \frac{1}{120} + \frac{k}{50}$ seconds.

💡 Solve for the angle first, then divide by $\omega$ to get time.

5. Applications: Finding specific times when an AC voltage reaches a peak or threshold

Applications: Finding Specific Times When an AC Voltage Reaches a Peak or Threshold

In electrical engineering, AC voltage is modeled as $V(t) = V_0 \sin(\omega t + \phi)$ or $V_0 \cos(\omega t + \phi)$ , where $V_0$ is peak voltage, $\omega$ is angular frequency, and $\phi$ is phase shift. Solving trigonometric equations determines when voltage reaches specific values.

Intuition: Finding when $V(t) = V_{\text{threshold}}$ translates to solving a basic trigonometric equation for $t$ within a physical time interval.

Core Rules:

Set $V_0 \sin(\omega t + \phi) = V_{\text{threshold}}$ and isolate the trigonometric function.
Solve for the argument $\omega t + \phi$ using general solution templates.
Solve for $t$ and restrict to the relevant time domain (e.g., $t \geq 0$ , one cycle $[0, \frac{2\pi}{\omega})$ ).
Interpret solutions in context (e.g., first occurrence, all occurrences in one period).

Consequence: This method identifies critical moments for circuit design, safety thresholds, and signal analysis.

Example: For $V(t) = 120\sin(120\pi t)$ volts, find when $V = 60$ in the first cycle. Solve $120\sin(120\pi t) = 60$ , so $\sin(120\pi t) = \frac{1}{2}$ , giving $t = \frac{1}{720}$ s and $t = \frac{5}{720}$ s.

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Level 3

MOD: TRANSLATE

The AC voltage of a machine is modeled by the equation $V(t) = 100 \sin(100\pi t)$ . Find the first time $t > 0$ (in seconds) when the voltage reaches $50$ volts.

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Basic trigonometric equations

✖️ 1. General solution templates for sine, cosine, and tangent

🔄 General Solution Templates

1. General solution templates for sine, cosine, and tangent

General Solution Templates for Sine, Cosine, and Tangent

✖️ 2. Solving equations within a restricted domain

📏 Restricted Domain Solutions

2. Solving equations within a restricted domain

Solving Equations Within a Restricted Domain

✖️ 3. Factoring trigonometric expressions and solving via the zero-product property

✂️ Factoring and Zero-Product Property

3. Factoring trigonometric expressions and solving via the zero-product property

Factoring Trigonometric Expressions and Solving via the Zero-Product Property

✖️ 4. Quadratic substitution methods and checking for extraneous solutions

🔁 Quadratic Substitution (u-substitution)

4. Quadratic substitution methods and checking for extraneous solutions

Quadratic Substitution Methods and Checking for Extraneous Solutions

✖️ 5. Applications: Finding specific times when an AC voltage reaches a peak or threshold

⚡ AC Voltage Timing Problems

5. Applications: Finding specific times when an AC voltage reaches a peak or threshold

Applications: Finding Specific Times When an AC Voltage Reaches a Peak or Threshold

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