Exercises — Solving trig equations — general solutions, solutions in given range

WHAT: this is a cosine equation, so we use the $2n\pi \pm \alpha$ machine (in degrees, $360^{\circ }n\pm \alpha$). WHY $\pm$: cosine is even, $\cos (-\theta )=\cos \theta$ — reflecting across the $x$-axis (look at the two red dots in the figure, mirror images top and bottom) gives a second solution with the same $x$-coordinate.

• Principal value: $\alpha ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=30^{\circ }$.
• General: $\theta =360^{\circ }n\pm 30^{\circ }$.
• $n=0$: $\theta =30^{\circ }$ ✓, and $-30^{\circ }$ (out of range).
• $n=1$: $\theta =360^{\circ }-30^{\circ }=330^{\circ }$ ✓ (and $390^{\circ }$ out).
• Answers: $\theta =30^{\circ },330^{\circ }$.

WHAT: sine equation → use $n\pi +(-1{)}^n\alpha$.

• $\alpha ={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$.
• $n=0$: $\theta =\frac{\pi }{6}$ ✓.
• $n=1$: $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ ✓ (the $(-1{)}^1=-1$ flips the sign, then $+\pi$ shifts — this is the supplementary angle).
• $n=2$: $\theta =2\pi +\frac{\pi }{6}$ (out of range, stop).
• Answers: $\theta =\frac{\pi }{6},\frac{5\pi }{6}$.

WHAT: tangent → use $180^{\circ }n+\alpha$. One family only, because $\tan$ repeats every $180^{\circ }$ and hits each value once per repeat.

• $\alpha ={\tan }^{-1}(\sqrt{3})=60^{\circ }$.
• $n=0$: $60^{\circ }$ ✓.
• $n=1$: $60^{\circ }+180^{\circ }=240^{\circ }$ ✓.
• $n=2$: $420^{\circ }$ (out).
• Answers: $\theta =60^{\circ },240^{\circ }$.

WHY the negative matters: a negative cosine means the $x$-coordinate is negative → the point sits in the left half of the circle (quadrants II and III). The inverse-cosine key handles this for us by returning an obtuse principal value.

• $\alpha ={\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi }{3}$.
• General: $\theta =2n\pi \pm \frac{2\pi }{3}$.
• $n=0$: $\theta =\frac{2\pi }{3}$ ✓, and $-\frac{2\pi }{3}$ (out).
• $n=1$: $\theta =2\pi -\frac{2\pi }{3}=\frac{4\pi }{3}$ ✓.
• Answers: $\theta =\frac{2\pi }{3},\frac{4\pi }{3}$.

WHY negative: $\sin$ negative → point in the lower half (quadrants III, IV). The calculator gives a negative principal value; the formula sorts out placing it in range.

• $\alpha ={\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-60^{\circ }$.
• General: $\theta =180^{\circ }n+(-1{)}^n(-60^{\circ })$.
• $n=0$: $\theta =-60^{\circ }$ (out).
• $n=1$: $\theta =180^{\circ }-(-60^{\circ })=240^{\circ }$ ✓.
• $n=2$: $\theta =360^{\circ }+(-60^{\circ })=300^{\circ }$ ✓.
• $n=3$: $\theta =540^{\circ }+60^{\circ }$ (out).
• Answers: $\theta =240^{\circ },300^{\circ }$.

WHAT / WHY substitute: let $\varphi =\theta -30^{\circ }$. Solving the inside first is cleaner; we adjust the range to match.

• New range: $0^{\circ }\le \theta <360^{\circ }\Rightarrow -30^{\circ }\le \varphi <330^{\circ }$.
• $\alpha ={\tan }^{-1}(1)=45^{\circ }$. General: $\varphi =180^{\circ }n+45^{\circ }$.
• $n=0$: $\varphi =45^{\circ }$ ✓. $n=1$: $\varphi =225^{\circ }$ ✓. $n=-1$: $\varphi =-135^{\circ }$ (out). $n=2$: $405^{\circ }$ (out).
• Back-substitute $\theta =\varphi +30^{\circ }$: $\theta =75^{\circ },255^{\circ }$.
• Answers: $\theta =75^{\circ },255^{\circ }$.

WHY widen: let $\varphi =2\theta$. As $\theta$ runs over $[0,2\pi )$, $\varphi =2\theta$ runs over $[0,4\pi )$ — twice as far. Solve in the wide range, then divide by $2$. The picture below shows the doubling: one lap of $\theta$ becomes two laps of $\varphi$.

• $\alpha ={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$. General: $\varphi =n\pi +(-1{)}^n\frac{\pi }{6}$.
• $n=0$: $\varphi =\frac{\pi }{6}$.
• $n=1$: $\varphi =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.
• $n=2$: $\varphi =2\pi +\frac{\pi }{6}=\frac{13\pi }{6}$.
• $n=3$: $\varphi =3\pi -\frac{\pi }{6}=\frac{17\pi }{6}$.
• $n=4$: $\varphi =4\pi +\frac{\pi }{6}$ (out, $\ge 4\pi$).
• All four are $<4\pi$, so keep all four. Divide by $2$:
• Answers: $\theta =\frac{\pi }{12},\frac{5\pi }{12},\frac{13\pi }{12},\frac{17\pi }{12}$.

• Let $\varphi =3\theta$. Range widens $\times 3$: $0^{\circ }\le \varphi <540^{\circ }$.
• $\alpha ={\cos }^{-1}(0)=90^{\circ }$. General: $\varphi =360^{\circ }n\pm 90^{\circ }$.
• $n=0$: $\varphi =90^{\circ }$ ✓; $-90^{\circ }$ (out).
• $n=1$: $\varphi =360^{\circ }\pm 90^{\circ }=270^{\circ },450^{\circ }$ ✓ (both $<540^{\circ }$).
• $n=2$: $\varphi =720^{\circ }\pm 90^{\circ }$ (both out).
• $\varphi =90^{\circ },270^{\circ },450^{\circ }$. Divide by $3$:
• Answers: $\theta =30^{\circ },90^{\circ },150^{\circ }$.

WHY treat as a quadratic: every term is in $\cos \theta$. Let $c=\cos \theta$; then it's $2c^2-c-1=0$, a quadratic we can factor.

• Factor: $(2c+1)(c-1)=0\Rightarrow c=-\frac{1}{2}$ or $c=1$.
• Branch $\cos \theta =1$: $\alpha =0^{\circ }$, general $360^{\circ }n$. In range: $\theta =0^{\circ }$.
• Branch $\cos \theta =-\frac{1}{2}$: $\alpha =120^{\circ }$, general $360^{\circ }n\pm 120^{\circ }$. In range: $\theta =120^{\circ },240^{\circ }$.
• Answers: $\theta =0^{\circ },120^{\circ },240^{\circ }$.

WHY an identity: the two terms use different angles ($2\theta$ and $\theta$). Use the double-angle form $\cos 2\theta =1-2{\sin }^2\theta$ (Compound and Double Angle Formulae) so everything is in $\sin \theta$.

• Substitute: $1-2{\sin }^2\theta +\sin \theta =0\Rightarrow 2{\sin }^2\theta -\sin \theta -1=0$.
• Let $s=\sin \theta$: $2s^2-s-1=0\Rightarrow (2s+1)(s-1)=0\Rightarrow s=-\frac{1}{2}$ or $s=1$.
• Branch $\sin \theta =1$: $\theta =\frac{\pi }{2}$ (only solution in range; it's a peak, one point).
• Branch $\sin \theta =-\frac{1}{2}$: $\alpha =-\frac{\pi }{6}$, general $n\pi +(-1{)}^n(-\frac{\pi }{6})$.
  • $n=1$: $\pi +\frac{\pi }{6}=\frac{7\pi }{6}$ ✓.
  • $n=2$: $2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$ ✓.
• Answers: $\theta =\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$.

WHY rewrite: $\cot \theta =\frac{1}{\tan \theta }$, so with $t=\tan \theta$ the equation is $t+\frac{1}{t}=2$. Multiply by $t$ (allowed: $\tan \theta =0$ never satisfies the original since $\cot$ would blow up):

• $t^2-2t+1=0\Rightarrow (t-1{)}^2=0\Rightarrow t=1$.
• $\tan \theta =1\Rightarrow \alpha =\frac{\pi }{4}$, general $n\pi +\frac{\pi }{4}$.
• $n=0$: $\frac{\pi }{4}$ ✓. $n=1$: $\frac{5\pi }{4}$ ✓. $n=2$: $\frac{9\pi }{4}$ (out).
• Answers: $\theta =\frac{\pi }{4},\frac{5\pi }{4}$.

WHY the double-angle expansion: $\sin 2\theta =2\sin \theta \cos \theta$ (Compound and Double Angle Formulae) brings both sides to a common footing so we can factor.

• $2\sin \theta \cos \theta -\cos \theta =0\Rightarrow \cos \theta (2\sin \theta -1)=0$.
• Branch $\cos \theta =0$: $\theta =\frac{\pi }{2},\frac{3\pi }{2}$.
• Branch $\sin \theta =\frac{1}{2}$: $\alpha =\frac{\pi }{6}$ → $\theta =\frac{\pi }{6},\frac{5\pi }{6}$.
• Answers: $\theta =\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{3\pi }{2}$.

WHY this identity: RHS is in $\sin$, so pick $\cos 2\theta =1-2{\sin }^2\theta$ to make the whole thing a quadratic in $\sin \theta$.

• $1-2{\sin }^2\theta =\sin \theta \Rightarrow 2{\sin }^2\theta +\sin \theta -1=0$.
• Factor: $(2\sin \theta -1)(\sin \theta +1)=0\Rightarrow \sin \theta =\frac{1}{2}$ or $\sin \theta =-1$.
• Branch $\sin \theta =\frac{1}{2}$: general $n\pi +(-1{)}^n\frac{\pi }{6}$.
  • $n=0$: $\frac{\pi }{6}$ ✓. $n=1$: $\frac{5\pi }{6}$ ✓. Negative-side $n=-1$: $-\pi -\frac{\pi }{6}$ (out). Both kept values lie in $[-\pi ,\pi ]$.
• Branch $\sin \theta =-1$: trough at $\theta =-\frac{\pi }{2}$ (in range; $\frac{3\pi }{2}$ out of range).
• Answers: $\theta =-\frac{\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$.

$\cos \theta =c$ uses which general solution? ::: $\theta =2n\pi \pm \alpha$ $\sin \theta =s$ uses which general solution? ::: $\theta =n\pi +(-1{)}^n\alpha$ $\tan \theta =t$ uses which general solution? ::: $\theta =n\pi +\alpha$ For $\sin 2\theta =c$ over $[0,2\pi )$, what range do you solve the inside over? ::: $[0,4\pi )$ When the range includes negatives, what must you remember to try? ::: negative values of $n$