This page presents effective methods for solving trigonometric equations, featuring detailed step-by-step solutions and graphical interpretations to enhance understanding of trigonometric identities and unit circle properties.
Step-by-Step Practice Problems
Problem 1: Basic Sine Equation
Find all solutions of the equation \(\sin(x) = \frac{1}{2}\) in the interval \([0 , 2\pi)\) and explain the solutions graphically.
Solution:
\(\sin(x)\) is positive in quadrants I and II. In Q1, the solution is \(x = \frac{\pi}{6}\). In Q2, \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
Problem 2: Linear Trig Equation
Find all solutions of the equation \(\sin(x) + \cos(x) = 0\) in \([0 , 2\pi)\).
Solution:
\(\sin(x) = -\cos(x) \implies \frac{\sin(x)}{\cos(x)} = -1 \implies \tan(x) = -1\).
\(\tan(x)\) is negative in Q2 and Q4. Solutions are \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\).
Problem 3: Compound Angles
Solve: \( \sin\left(\frac{\pi x}{2}\right)\cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi x}{2}\right)\sin\left(\frac{\pi}{3}\right) = 0 \).
Solution:
Using \(\sin(A-B) = \sin A \cos B - \cos A \sin B\), the equation simplifies to \(\sin\left(\frac{\pi x}{2} - \frac{\pi}{3}\right) = 0\).
\(\frac{\pi x}{2} - \frac{\pi}{3} = k\pi \implies x = 2k + \frac{2}{3}\).
Problem 4: Quadratic Form (Sine)
Solve the equation: \(2 \sin^2(x) + \cos(x) = 1\).
Solution:
Using \(\sin^2(x) = 1 - \cos^2(x)\), we get \(2(1-\cos^2(x)) + \cos(x) = 1 \implies -2\cos^2(x) + \cos(x) + 1 = 0\).
Let \(u = \cos(x) \implies -2u^2 + u + 1 = 0 \implies (u-1)(-2u-1) = 0\).
\(\cos(x) = 1 \implies x = 0, 2\pi\); \(\cos(x) = -1/2 \implies x = 2\pi/3, 4\pi/3\).
Problem 5: Half Angle Form
Solve: \(6 \cos^2\left(\frac{x}{2}\right) - \cos(x) = 4\).
Solution:
Use \(\cos(x) = 2\cos^2(x/2) - 1\). The equation becomes \(4\cos^2(x/2) = 3 \implies \cos(x/2) = \pm\frac{\sqrt{3}}{2}\).
Solving for \(x\) yields \(x = \frac{\pi}{3} + 4k\pi, \frac{11\pi}{3} + 4k\pi, \frac{5\pi}{3} + 4k\pi, \frac{7\pi}{3} + 4k\pi\).
Problem 6: Secant & Tangent
Solve: \(\tan^2(x) + 2 \sec(x) + 1 = 0\) in \([0 , 2\pi)\) .
Solution:
Using \(\tan^2(x) = \sec^2(x) - 1\), we get \(\sec^2(x) + 2\sec(x) = 0 \implies \sec(x)(\sec(x)+2) = 0\).
\(\sec(x) = 0\) (impossible); \(\sec(x) = -2 \implies \cos(x) = -1/2\). Solutions: \(2\pi/3, 4\pi/3\).
Problem 7: Sum and Difference Identities
Solve: \(\cos(3x) + \sin(2x) = \cos(x)\).
Solution:
Rewrite using \(\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x)\) and \(\sin(2x) = 2\sin(x)\cos(x)\).
After factoring \(\cos(x)\), we obtain \(2\sin(x)\cos(x)(1 - 2\sin(x)) = 0\). Solutions include \(x = k\pi, \pi/2 + k\pi, \pi/6 + 2k\pi, 5\pi/6 + 2k\pi\).
Problem 8: Cosine Equality
Solve: \(\cos(3x) = \cos\left(2x + \frac{\pi}{4}\right)\).
Solution:
Apply the rule: if \(\cos(A) = \cos(B)\), then \(A = B + 2k\pi\) or \(A = -B + 2k\pi\).
\(3x = 2x + \frac{\pi}{4} + 2k\pi \implies x = \frac{\pi}{4} + 2k\pi\).
\(3x = -(2x + \frac{\pi}{4}) + 2k\pi \implies 5x = -\frac{\pi}{4} + 2k\pi \implies x = -\frac{\pi}{20} + \frac{2k\pi}{5}\).
Challenge Problems
Challenge 1: Quadratic in Sine
Solve: \(2\sin^2(x) + 3\cos(x) = 0\) for \(x \in [0, 2\pi]\).
Solution:
Use \(\sin^2(x) = 1 - \cos^2(x) \implies -2\cos^2(x) + 3\cos(x) + 2 = 0\).
Factor \((2\cos(x) + 1)(\cos(x) - 2) = 0\).
\(\cos(x) = -1/2 \implies x = 2\pi/3, 4\pi/3\). (\(\cos(x) = 2\) is impossible).
Challenge 2: Sum-to-Product Proof
Prove: \(\sin(3x) + \sin(x) = 2\sin(2x)\cos(x)\).
Solution:
Use \(\sin(A) + \sin(B) = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\).
\(2\sin\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) = 2\sin(2x)\cos(x)\). Identity proven.