5.4 Trigonometric equations
An equation like \(\sin \theta =\frac {1}{2}\) has infinitely many solutions, because the ratios repeat. A question therefore always states a range, and the task is to find every solution inside it.
Solution. Step 1 — the acute angle. \(\sin 30^\circ =\frac {1}{2}\), so the associated acute angle is \(30^\circ \).
Step 2 — which quadrants. Sine is positive, so \(\theta \) is in the first or second quadrant.
Step 3 — build the answers. \[\text {first quadrant: } \theta =30^\circ ,\qquad \text {second quadrant: } \theta =180^\circ -30^\circ =150^\circ .\] \[\therefore \quad \theta =30^\circ \text { or } 150^\circ .\]
A calculator gives only \(30^\circ \). The second solution has to be reasoned out, and forgetting it is the standard way to lose half the marks.
Solution. Treat it as a quadratic in \(\sin \theta \). Writing \(s=\sin \theta \): \[2s^2+s-1=0\quad \implies \quad (2s-1)(s+1)=0\] \[\implies \quad s=\frac {1}{2}\quad \text {or}\quad s=-1.\]
Case \(\sin \theta =\frac {1}{2}\). From the previous example, \(\theta =30^\circ \) or \(150^\circ \).
Case \(\sin \theta =-1\). Sine reaches \(-1\) only at the bottom of the circle: \(\theta =270^\circ \).
\[\therefore \quad \theta =30^\circ ,\ 150^\circ \text { or } 270^\circ .\]
Check. At \(\theta =270^\circ \): \(2(-1)^2+(-1)-1=2-1-1=0\). \(\relax \amscheckmark \)
Note 5.16. \(\sin \theta =-1\) gives one solution, not two, because it occurs at the boundary between quadrants rather than inside one. The same is true of \(\sin \theta =1\), \(\cos \theta =\pm 1\) and \(\sin \theta =0\). Applying the “two quadrants” rule blindly here produces a duplicate answer.
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