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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 2cosโก2xโˆ’cosโกxโˆ’1=02\cos^2 x - \cos x - 1 = 0 on the interval [0,2ฯ€)[0, 2\pi).

  1. Treat the trigonometric function as a variable. Imagine u=cosโกxu = \cos x. The equation becomes 2u2โˆ’uโˆ’1=02u^2 - u - 1 = 0.
  2. Factor the quadratic. (2u+1)(uโˆ’1)=0(2u + 1)(u - 1) = 0.
  3. Substitute the trig function back in. (2cosโกx+1)(cosโกxโˆ’1)=0(2\cos x + 1)(\cos x - 1) = 0.
  4. Set each factor to zero.
    • 2cosโกx+1=0โ€…โ€ŠโŸนโ€…โ€Šcosโกx=โˆ’122\cos x + 1 = 0 \implies \cos x = -\frac{1}{2}
    • cosโกxโˆ’1=0โ€…โ€ŠโŸนโ€…โ€Šcosโกx=1\cos x - 1 = 0 \implies \cos x = 1
  5. Find the angles on [0,2ฯ€)[0, 2\pi).
    • For cosโกx=โˆ’12\cos x = -\frac{1}{2}, the solutions are x=2ฯ€3x = \frac{2\pi}{3} and x=4ฯ€3x = \frac{4\pi}{3}.
    • For cosโกx=1\cos x = 1, the solution is x=0x = 0.

The final solutions are x=0,2ฯ€3,4ฯ€3x = 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 sinโก2x=cosโกx\sin 2x = \cos x on the interval [0,2ฯ€)[0, 2\pi).

  1. Use the double-angle identity. We know that sinโก2x=2sinโกxcosโกx\sin 2x = 2\sin x \cos x.
  2. Rewrite the equation. 2sinโกxcosโกx=cosโกx2\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! 2sinโกxcosโกxโˆ’cosโกx=02\sin x \cos x - \cos x = 0 cosโกx(2sinโกxโˆ’1)=0\cos x(2\sin x - 1) = 0
  4. Set each factor to zero.
    • cosโกx=0โ€…โ€ŠโŸนโ€…โ€Šx=ฯ€2,3ฯ€2\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}
    • 2sinโกxโˆ’1=0โ€…โ€ŠโŸนโ€…โ€Šsinโกx=12โ€…โ€ŠโŸนโ€…โ€Šx=ฯ€6,5ฯ€62\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}

The final solutions are x=ฯ€6,ฯ€2,5ฯ€6,3ฯ€2x = \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 sinโกx\sin x on both sides, subtract it to one side and factor it out. Dividing by sinโกx\sin x throws away the solutions where sinโกx=0\sin x = 0.
  • Check your interval: Pay close attention to the domain (e.g., [0,2ฯ€)[0, 2\pi) vs (โˆ’โˆž,โˆž)(-\infty, \infty)). If no interval is given, you must write a general solution by adding +2ฯ€k+ 2\pi k or +ฯ€k+ \pi k to your answers.