Solving Trigonometric Equations

Solve trigonometric equations using algebraic techniques, inverse functions, and the unit circle.

Solving Trigonometric Equations

Basic Strategies

When solving trigonometric equations, we use several key techniques:

  1. Isolate the trigonometric function
  2. Use inverse trig functions or the unit circle
  3. Find all solutions in the given interval
  4. Consider periodicity for general solutions

Types of Trigonometric Equations

Type 1: Linear Equations

Example form: sin⁡(x)=a\sin(x) = a where −1≤a≤1-1 \leq a \leq 1

Solution method:

  1. Find the reference angle: θr=arcsin⁡(∣a∣)\theta_r = \arcsin(|a|)
  2. Determine which quadrants based on the sign of aa
  3. Find all solutions in [0,2π)[0, 2\pi)
  4. Add 2πn2\pi n for general solution

General solution patterns:

  • If sin⁡(x)=a\sin(x) = a: x=arcsin⁡(a)+2πnx = \arcsin(a) + 2\pi n or x=π−arcsin⁡(a)+2πnx = \pi - \arcsin(a) + 2\pi n
  • If cos⁡(x)=a\cos(x) = a: x=arccos⁡(a)+2πnx = \arccos(a) + 2\pi n or x=−arccos⁡(a)+2πnx = -\arccos(a) + 2\pi n
  • If tan⁡(x)=a\tan(x) = a: x=arctan⁡(a)+πnx = \arctan(a) + \pi n

Type 2: Quadratic Equations

Example form: sin⁡2(x)+bsin⁡(x)+c=0\sin^2(x) + b\sin(x) + c = 0

Solution method:

  1. Substitute u=sin⁡(x)u = \sin(x) (or the relevant trig function)
  2. Solve the quadratic equation for uu
  3. Solve sin⁡(x)=u\sin(x) = u for each value of uu
  4. Check that solutions are in the domain

Type 3: Equations with Multiple Angles

Example form: sin⁡(2x)=12\sin(2x) = \frac{1}{2}

Solution method:

  1. Let u=2xu = 2x (or the multiple angle)
  2. Solve sin⁡(u)=12\sin(u) = \frac{1}{2}
  3. Divide by the coefficient to find xx
  4. Consider the extended period

Type 4: Equations with Multiple Functions

Example form: sin⁡(x)+cos⁡(x)=1\sin(x) + \cos(x) = 1

Solution methods:

  • Use Pythagorean identities to express in terms of one function
  • Square both sides (check for extraneous solutions)
  • Use sum-to-product or product-to-sum identities

Key Identities for Solving

Pythagorean Identities

  • sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1
  • 1+tan⁡2(x)=sec⁡2(x)1 + \tan^2(x) = \sec^2(x)
  • 1+cot⁡2(x)=csc⁡2(x)1 + \cot^2(x) = \csc^2(x)

Double Angle Formulas

  • sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)
  • cos⁡(2x)=cos⁡2(x)−sin⁡2(x)=2cos⁡2(x)−1=1−2sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)
  • tan⁡(2x)=2tan⁡(x)1−tan⁡2(x)\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}

Half Angle Formulas

  • sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1 - \cos(2x)}{2}
  • cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1 + \cos(2x)}{2}

Interval Considerations

  • Standard interval: [0,2π)[0, 2\pi) or [0°,360°)[0°, 360°)
  • Extended intervals: May need to consider multiple periods
  • General solution: Include +2πn+ 2\pi n (or +πn+ \pi n for tangent) where nn is an integer

Checking Solutions

Always verify solutions by:

  1. Substituting back into the original equation
  2. Checking domain restrictions (e.g., no division by zero)
  3. Eliminating extraneous solutions from squaring

Common Reference Angles

| Angle | 0°0° | 30°30° | 45°45° | 60°60° | 90°90° | |-------|------|-------|-------|-------|-------| | Radians | 00 | π6\frac{\pi}{6} | π4\frac{\pi}{4} | π3\frac{\pi}{3} | π2\frac{\pi}{2} | | sin⁡\sin | 00 | 12\frac{1}{2} | 22\frac{\sqrt{2}}{2} | 32\frac{\sqrt{3}}{2} | 11 | | cos⁡\cos | 11 | 32\frac{\sqrt{3}}{2} | 22\frac{\sqrt{2}}{2} | 12\frac{1}{2} | 00 | | tan⁡\tan | 00 | 33\frac{\sqrt{3}}{3} | 11 | 3\sqrt{3} | undefined |

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve 2sin⁡(x)−1=02\sin(x) - 1 = 0 for xx in [0,2π)[0, 2\pi).

💡 Show Solution

Solution:

Starting equation: 2sin⁡(x)−1=02\sin(x) - 1 = 0

Step 1: Isolate the trig function 2sin⁡(x)=12\sin(x) = 1 sin⁡(x)=12\sin(x) = \frac{1}{2}

Step 2: Find solutions using the unit circle

sin⁡(x)=12\sin(x) = \frac{1}{2} when xx is in Quadrants I and II (where sine is positive).

Reference angle: arcsin⁡(12)=π6\arcsin(\frac{1}{2}) = \frac{\pi}{6}

Step 3: Find all solutions in [0,2π)[0, 2\pi)

  • Quadrant I: x=π6x = \frac{\pi}{6}
  • Quadrant II: x=π−π6=5π6x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

Answer: x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}

Verification:

  • 2sin⁡(π6)−1=2(12)−1=02\sin(\frac{\pi}{6}) - 1 = 2(\frac{1}{2}) - 1 = 0 ✓
  • 2sin⁡(5π6)−1=2(12)−1=02\sin(\frac{5\pi}{6}) - 1 = 2(\frac{1}{2}) - 1 = 0 ✓

2Problem 2medium

❓ Question:

Solve for xx in the interval [0,2π)[0, 2\pi):

a) 2sin⁡x=32\sin x = \sqrt{3} b) cos⁡2x=14\cos^2 x = \frac{1}{4} c) 2cos⁡x−1=02\cos x - 1 = 0

💡 Show Solution

Solution:

Part (a): 2sin⁡x=32\sin x = \sqrt{3}

sin⁡x=32\sin x = \frac{\sqrt{3}}{2}

This occurs at x=π3x = \frac{\pi}{3} (Quadrant I) and x=2π3x = \frac{2\pi}{3} (Quadrant II)

Solutions: x=π3,2π3x = \frac{\pi}{3}, \frac{2\pi}{3}

Part (b): cos⁡2x=14\cos^2 x = \frac{1}{4}

cos⁡x=±12\cos x = \pm\frac{1}{2}

For cos⁡x=12\cos x = \frac{1}{2}: x=π3,5π3x = \frac{\pi}{3}, \frac{5\pi}{3}

For cos⁡x=−12\cos x = -\frac{1}{2}: x=2π3,4π3x = \frac{2\pi}{3}, \frac{4\pi}{3}

Solutions: x=π3,2π3,4π3,5π3x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}

Part (c): 2cos⁡x−1=02\cos x - 1 = 0

cos⁡x=12\cos x = \frac{1}{2}

Solutions: x=π3,5π3x = \frac{\pi}{3}, \frac{5\pi}{3}

3Problem 3medium

❓ Question:

Solve 2cos⁡2(x)−cos⁡(x)−1=02\cos^2(x) - \cos(x) - 1 = 0 for xx in [0,2π)[0, 2\pi).

💡 Show Solution

Solution:

Given: 2cos⁡2(x)−cos⁡(x)−1=02\cos^2(x) - \cos(x) - 1 = 0

This is a quadratic equation in cos⁡(x)\cos(x).

Step 1: Factor or use quadratic formula

Let u=cos⁡(x)u = \cos(x): 2u2−u−1=02u^2 - u - 1 = 0

Factor: (2u+1)(u−1)=0(2u + 1)(u - 1) = 0

Step 2: Solve for uu 2u+1=0oru−1=02u + 1 = 0 \quad \text{or} \quad u - 1 = 0 u=−12oru=1u = -\frac{1}{2} \quad \text{or} \quad u = 1

Step 3: Solve for xx

Case 1: cos⁡(x)=−12\cos(x) = -\frac{1}{2}

Cosine is negative in Quadrants II and III. Reference angle: arccos⁡(12)=π3\arccos(\frac{1}{2}) = \frac{\pi}{3}

  • Quadrant II: x=π−π3=2π3x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
  • Quadrant III: x=π+π3=4π3x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

Case 2: cos⁡(x)=1\cos(x) = 1

x=0x = 0 (or 2π2\pi, but we use [0,2π)[0, 2\pi))

Answer: x=0,2π3,4π3x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}

Verification:

  • At x=0x = 0: 2(1)2−1−1=02(1)^2 - 1 - 1 = 0 ✓
  • At x=2π3x = \frac{2\pi}{3}: 2(−12)2−(−12)−1=12+12−1=02(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0 ✓
  • At x=4π3x = \frac{4\pi}{3}: 2(−12)2−(−12)−1=02(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1 = 0 ✓

4Problem 4hard

❓ Question:

Solve for xx in [0,2π)[0, 2\pi):

sin⁡(2x)=cos⁡x\sin(2x) = \cos x

💡 Show Solution

Solution:

Use the double-angle formula: sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x

2sin⁡xcos⁡x=cos⁡x2\sin x \cos x = \cos x

2sin⁡xcos⁡x−cos⁡x=02\sin x \cos x - \cos x = 0

cos⁡x(2sin⁡x−1)=0\cos x(2\sin x - 1) = 0

This gives us two cases:

Case 1: cos⁡x=0\cos x = 0

In [0,2π)[0, 2\pi): x=π2,3π2x = \frac{\pi}{2}, \frac{3\pi}{2}

Case 2: 2sin⁡x−1=02\sin x - 1 = 0

sin⁡x=12\sin x = \frac{1}{2}

In [0,2π)[0, 2\pi): x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}

All solutions: x=π6,π2,5π6,3π2x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}

5Problem 5hard

❓ Question:

Solve sin⁡(2x)=sin⁡(x)\sin(2x) = \sin(x) for xx in [0,2π)[0, 2\pi).

💡 Show Solution

Solution:

Given: sin⁡(2x)=sin⁡(x)\sin(2x) = \sin(x)

Step 1: Use double angle formula 2sin⁡(x)cos⁡(x)=sin⁡(x)2\sin(x)\cos(x) = \sin(x)

Step 2: Move all terms to one side 2sin⁡(x)cos⁡(x)−sin⁡(x)=02\sin(x)\cos(x) - \sin(x) = 0

Step 3: Factor out sin⁡(x)\sin(x) sin⁡(x)(2cos⁡(x)−1)=0\sin(x)(2\cos(x) - 1) = 0

Step 4: Solve each factor

Factor 1: sin⁡(x)=0\sin(x) = 0

In [0,2π)[0, 2\pi): x=0,πx = 0, \pi

Factor 2: 2cos⁡(x)−1=02\cos(x) - 1 = 0 cos⁡(x)=12\cos(x) = \frac{1}{2}

Cosine is positive in Quadrants I and IV. Reference angle: arccos⁡(12)=π3\arccos(\frac{1}{2}) = \frac{\pi}{3}

  • Quadrant I: x=π3x = \frac{\pi}{3}
  • Quadrant IV: x=2π−π3=5π3x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

Answer: x=0,π3,π,5π3x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}

Verification:

  • At x=0x = 0: sin⁡(0)=sin⁡(0)=0\sin(0) = \sin(0) = 0 ✓
  • At x=π3x = \frac{\pi}{3}: sin⁡(2π3)=32\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} and sin⁡(π3)=32\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} ✓
  • At x=πx = \pi: sin⁡(2π)=sin⁡(π)=0\sin(2\pi) = \sin(\pi) = 0 ✓
  • At x=5π3x = \frac{5\pi}{3}: sin⁡(10π3)=sin⁡(4π3)=−32\sin(\frac{10\pi}{3}) = \sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2} and sin⁡(5π3)=−32\sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2} ✓

Note: We reduced sin⁡(10π3)\sin(\frac{10\pi}{3}) by subtracting 2π2\pi: 10π3−2π=4π3\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}

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