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Solving Trigonometric Equations

Solving Trigonometric Equations

Solving trigonometric equations involves finding the unknown angle(s) that make the equation true. In advanced trigonometry, these equations often look like algebraic equations (such as quadratics) or contain multiple different angles, requiring you to use trigonometric identities to simplify them first.

Quadratic-Form Trigonometric Equations

Sometimes, a trigonometric equation takes the form of a quadratic equation. You can solve these by factoring, just as you would with a regular polynomial.

Example: Solve 2cos⁡2x−cos⁡x−1=02\cos^2 x - \cos x - 1 = 0 on the interval [0,2π)[0, 2\pi).

  1. Treat the trigonometric function as a variable. Imagine u=cos⁡xu = \cos x. The equation becomes 2u2−u−1=02u^2 - u - 1 = 0.
  2. Factor the quadratic. (2u+1)(u−1)=0(2u + 1)(u - 1) = 0.
  3. Substitute the trig function back in. (2cos⁡x+1)(cos⁡x−1)=0(2\cos x + 1)(\cos x - 1) = 0.
  4. Set each factor to zero.
    • 2cos⁡x+1=0  ⟹  cos⁡x=−122\cos x + 1 = 0 \implies \cos x = -\frac{1}{2}
    • cos⁡x−1=0  ⟹  cos⁡x=1\cos x - 1 = 0 \implies \cos x = 1
  5. Find the angles on [0,2π)[0, 2\pi).
    • For cos⁡x=−12\cos x = -\frac{1}{2}, the solutions are x=2π3x = \frac{2\pi}{3} and x=4π3x = \frac{4\pi}{3}.
    • For cos⁡x=1\cos x = 1, the solution is x=0x = 0.

The final solutions are x=0,2π3,4π3x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}.

Equations with Multiple Angles

If an equation contains different arguments (like 2x2x and xx), you must use trigonometric identities to rewrite the equation so that all trig functions have the same angle.

Example: Solve sin⁡2x=cos⁡x\sin 2x = \cos x on the interval [0,2π)[0, 2\pi).

  1. Use the double-angle identity. We know that sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x.
  2. Rewrite the equation. 2sin⁡xcos⁡x=cos⁡x2\sin x \cos x = \cos x.
  3. Move all terms to one side and factor. Never divide by a variable trig function, or you will lose valid solutions! 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
  4. Set each factor to zero.
    • cos⁡x=0  ⟹  x=π2,3π2\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}
    • 2sin⁡x−1=0  ⟹  sin⁡x=12  ⟹  x=π6,5π62\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}

The final solutions are x=π6,π2,5π6,3π2x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}.

Key Strategies for Success

  • Know your identities: Pythagorean, double-angle, and reciprocal identities are your primary tools for rewriting equations.
  • Factor, don't divide: If you have sin⁡x\sin x on both sides, subtract it to one side and factor it out. Dividing by sin⁡x\sin x throws away the solutions where sin⁡x=0\sin x = 0.
  • Check your interval: Pay close attention to the domain (e.g., [0,2π)[0, 2\pi) vs (−∞,∞)(-\infty, \infty)). If no interval is given, you must write a general solution by adding +2πk+ 2\pi k or +πk+ \pi k to your answers.