Recipe

Solve a trig equation that needs an identity first

Use when: a trig equation has more than one function name, or the same function at two different powers, so the reference-angle method from the previous section doesn't apply directly

Goal: every solution in the given interval (or the general solution), found only after the equation has been rebuilt into single-function form

Taught in Equations requiring identities

1. 1

   Check for two symptoms: more than one function name in the equation, or the same function appearing at more than one power. Either symptom means the reference-angle method can't be applied directly, yet.

   Example

   \sin\theta=\cos\theta has two function names. 2\sin^2\theta-\sin\theta-1=0 has one function name at two powers. Both need rebuilding before reference angles apply.

   • only one function name, at one power — stop here, the equation is already basic; use the previous section's method directly.
   • one or both symptoms are present — continue to Step 2.

   Why: applying reference angles to an equation that isn't single-function yet produces nonsense; naming the actual obstacle first is what tells you which identity to reach for next.

2. 2

   For two mixed functions, use a reciprocal or quotient identity (or divide through) to rewrite one function in terms of the other. For one function at two powers, use a Pythagorean identity to replace the higher power, when that's what's producing it.

   Example

   \sin\theta=\cos\theta: check \cos\theta=0 directly first (it doesn't solve the equation), then divide by \cos\theta to get \tan\theta=1.

   • two different functions, and dividing is confirmed safe (see the division-trap check below before dividing) — divide both sides by the appropriate function to collapse to one function.
   • two different functions, but dividing risks losing a solution, or the equation can be rearranged to equal zero — move everything to one side and factor instead.

   Why: this is the step that actually removes the obstacle identified in Step 1. Every later step is just finishing an equation now in ordinary, solvable shape.

3. 3

   If the equation is now linear in one function, solve directly. If it's quadratic in one function (after a substitution like u=\sin\theta), solve the quadratic and discard any root outside [-1,1] for a sine or cosine substitution.

   Example

   2u^2-u-1=0 factors as (2u+1)(u-1)=0, giving u=-\tfrac12 or u=1; both are inside [-1,1], so neither is discarded.

   • linear in one function — isolate it and move to Step 4.
   • quadratic in one function — substitute u, solve au^2+bu+c=0 by factoring or the quadratic formula, discard any root outside [-1,1] for sine or cosine (tangent needs no such filter), then move to Step 4 with each surviving root.

   Why: once the equation is down to one function, it's an ordinary algebra problem the equations chapter already covers. The identity's job was only to get here.

4. 4

   Solve (\text{function})=k for every angle in the interval, exactly as in the previous section: find the reference angle, then use the sign of k to pick the correct quadrants.

   Why: this is the same finishing move every basic trig equation uses. The identity work in Steps 2 and 3 exists only to arrive at this familiar shape.

5. 5

   Substituting an identity doesn't just rewrite an equation for free. Dividing by an expression that could be zero can silently drop a solution, and squaring or multiplying by an expression that could be zero can introduce a false one.

   • you divided by a function in Step 2 — confirm the case you divided away isn't itself a solution of the ORIGINAL equation, by testing it directly.
   • you substituted a Pythagorean identity or squared something in Step 2 or 3 — plug every candidate solution back into the ORIGINAL equation, not the rewritten one, before reporting it.

   Why: the rewritten equation only ever remembers the case that survives it; the original equation is the only reliable judge of what actually solves it.