Solving trig equations

If you read the title as trick equations, you are not wrong.

Procedure

  1. Use identities, especially Pythagorean and double angle identities, to convert multiple trig functions into only sin⁡\sin or cos⁡\cos or tan⁡\tan. See list of identities at SL and identities at HL.
  2. Get to a linear or quadratic equation in a single trig function
  3. Get to a point where you can simply take the inverse trig function
  4. When the argument is something like ω(x−h)\omega(x - h) with domain restriction a≤x≤b{a \leq x \leq b}, then rewrite using ω(a−h)≤ω(x−h)≤ω(b−h){\omega(a - h) \leq \omega(x - h) \leq \omega (b - h)}, possibly flip the signs when ω<0\omega < 0.
  5. Identify two solutions for ω(x−h)\omega(x - h) in different quadrants. Other solutions differ by 2nπ2n\pi, ie co-terminal with the two.
Example Solutions
sin⁡x=k\sin x = k sin⁡−1k+2nππ−sin⁡−1k+2nπ\displaystyle \sin^{-1} k + 2n\pi \\ \pi -\sin^{-1} k + 2n\pi
cos⁡x=k\cos x = k cos⁡−1k+2nπ−cos⁡−1k+2nπ\displaystyle \cos^{-1} k + 2n\pi \\ -\cos^{-1} k + 2n\pi
tan⁡x=k\tan x = k tan⁡−1k+nπ\displaystyle \tan^{-1} k + n\pi

The solution for sin⁡\sin comes from the HL identity of sin⁡(x)=sin⁡(π−x)\sin(x) = \sin (\pi - x).

The solution for cos⁡\cos comes from the HL identity of cos⁡(x)=cos⁡(−x)\cos(x) = \cos (-x).

The solution for tan⁡\tan comes from the fact that tangent is the slope of a line through the origin, and the angle about the origin is π\pi radians.

Example: Solve sin⁡(2(x+π3))=−12,0≤x≤2π\displaystyle \sin\left(2\left(x + \frac{\pi}{3}\right)\right) = -\frac{1}{2}, 0 \leq x \leq 2\pi

Let y=2(x+π3)\displaystyle y = 2\left(x + \frac{\pi}{3}\right).

2(0+π3)≤y≤2(2π+π3)2π3≤y≤14π34π6≤y≤28π3\begin{align*} 2\left(0+\frac\pi3\right) &\leq y && \leq 2\left(2\pi+\frac\pi3\right) \\ \frac{2\pi}{3} &\leq y && \leq \frac{14\pi}{3} \\ \frac{4\pi}{6} &\leq y && \leq \frac{28\pi}{3} \end{align*}

sin⁡−1(−12)=−π6\sin^{-1} \left(-\frac12\right) = -\frac\pi6. The other base solution for yy is π−(−π6)=7π6\pi - \left(-\frac\pi6\right) = \frac{7\pi}{6}. We need all the co-terminal angles in the new interval, by adding 2π=12π62\pi = \frac{12\pi}6.

This means −π6,11π6,23π6-\frac{\pi}{6}, \frac{11\pi}{6}, \frac{23\pi}{6} and also 7π6,19π6\frac{7\pi}6, \frac{19\pi}{6}. −π6-\frac{\pi}6 can be rejected as it is outside the new interval.

2(x+π3)=7π6,11π6,19π6,23π6x+4π12=7π12,11π12,19π12,23π12x=3π12,7π12,15π12,19π12■\begin{align*} 2\left(x + \frac{\pi}{3}\right) &= \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}{6}, \frac{23\pi}{6} \\ x + \frac{4\pi}{12}&= \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12} \\ x &= \frac{3\pi}{12}, \frac{7\pi}{12}, \frac{15\pi}{12}, \frac{19\pi}{12} \qed \end{align*}

Tip: Fractions don’t need to be simplified (but denominators must be rational).

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