Every technique from Year 1 — sketch, find the principal value, use symmetry — carries over unchanged. Only the units of the answer, and the range they're searched over, are different.
8 min read3 worked examples
Same method, new units
Solving $\sin\theta=k$, $\cos\theta=k$ or $\tan\theta=k$ in radians uses exactly the same strategy as in Year 1: sketch the graph, find the calculator's principal value, then use the curve's symmetry to find every other solution inside the given range. The only changes are cosmetic but easy to trip over — the range is now written using $\pi$ (e.g. $0\le\theta\le2\pi$ instead of $0°\le\theta\le360°$), and a calculator answer needs to come out in radian mode, not degree mode.
Radians and degrees are never mixed
If the question's range is given in radians, every solution must be given in radians too — check the calculator is set to radian mode before finding a principal value. Mixing an exact-π range with a degree-mode calculator answer is one of the most common marks lost in this topic.
Exact-value equations (built from the Year 1 exact values, now labelled with $\pi$) usually have exact $\pi$-form answers; other equations need a decimal principal value, and answers are typically left to 2 or 3 decimal places (radians don't split into a "nice" number of decimal places the way degree answers to 1 d.p. do).
Worked example 1
An exact-value sine equation
Solve $\sin\theta = \dfrac12$ for $0 \le \theta \le 2\pi$, giving exact answers in terms of $\pi$.
MethodSketch, find the principal value, then reflect in the peak
1
Sketch $y=\sin\theta$ for $0\le\theta\le2\pi$ together with the horizontal line $y=\dfrac12$.
The picture shows how many times the line crosses the curve before any calculation.
2
Principal value: $\theta=\sin^{-1}\left(\dfrac12\right)=\dfrac{\pi}{6}$ — one of the Year 1 exact values, now labelled in radians.
The calculator's first solution, in radian mode.
3
The sine curve is symmetric about its peak at $\theta=\dfrac{\pi}{2}$, so the second crossing is $\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$.
Reflect the principal value in the line $\theta=\pi/2$, exactly as with $\theta=90°$ before.
4
Both $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ lie in $[0,2\pi]$, and the sketch shows no further crossings.
Check against the range and the picture.
5
Solutions: $\theta=\dfrac{\pi}{6}$ and $\theta=\dfrac{5\pi}{6}$.
Final answers, left exact.
▪ solved
Worked example 2
A decimal cosine equation
Solve $\cos\theta = -0.6$ for $0 \le \theta \le 2\pi$, giving answers correct to 3 significant figures.
MethodSame sketch-and-reflect approach, now with a decimal principal value
1
Sketch $y=\cos\theta$ for $0\le\theta\le2\pi$ with the horizontal line $y=-0.6$, sitting below the axis.
Visualise the two crossings before computing either one.
2
Principal value: $\theta=\cos^{-1}(-0.6)\approx2.214$ radians (in radian mode).
The calculator handles the negative input directly.
3
The cosine curve is symmetric about $\theta=\pi$, so the second crossing is $2\pi-2.214\approx4.069$ radians.
Reflect in $\theta=\pi$, exactly as reflecting in $180°$ before.
4
Both values lie in $[0,2\pi]$, and the sketch shows no further crossings.
Check against the range.
5
Solutions: $\theta\approx2.21$ and $\theta\approx4.07$ (3 s.f.).
Final answers, rounded as requested.
▪ solved
Examiner's eye
Radian answers are almost always rounded to 2 or 3 significant figures or decimal places — never to "1 d.p." the way a degree answer might be. A radian value like $2.2$ is a much coarser rounding than $2.21$, and questions will specify exactly which is wanted.
Worked example 3
A transformed argument · when the angle is 2θ
Solve $\tan2\theta = \sqrt3$ for $0 \le \theta \le \pi$.
Here the variable inside the equation is $2\theta$, not $\theta$. The safe method is unchanged from Year 1: solve for $2\theta$ first, over a range doubled to match, then divide every solution by 2 at the very end.
MethodSolve for 2θ over the doubled range, then halve
1
Since $0\le\theta\le\pi$, the new variable satisfies $0\le2\theta\le2\pi$.
Multiply every part of the original range by 2, to match the substitution.
2
Principal value: $2\theta=\tan^{-1}(\sqrt3)=\dfrac{\pi}{3}$.
An exact Year 1 value, in radians.
3
Tan has period $\pi$, so the next solution is $\dfrac{\pi}{3}+\pi=\dfrac{4\pi}{3}$; adding $\pi$ again gives $\dfrac{7\pi}{3}$, which is outside $[0,2\pi]$.
Add the period repeatedly until leaving the doubled range.
4
So $2\theta=\dfrac{\pi}{3}$ or $2\theta=\dfrac{4\pi}{3}$, giving $\theta=\dfrac{\pi}{6}$ or $\theta=\dfrac{2\pi}{3}$.
Divide every solution for 2θ by 2, only at the very last step.
5
Both $\dfrac{\pi}{6}$ and $\dfrac{2\pi}{3}$ lie in the original range $[0,\pi]$.
Final check against the range actually asked for, in θ.
▪ solved
Why double the range at all
Dividing the range only at the end (rather than dividing it first) is what guarantees no solutions for $2\theta$ get missed. Halving the range too early is one of the most common ways to lose a valid solution in this question type.
Common mistake
Leaving the calculator in degree mode while a question's range is written in radians — the principal value comes out numerically wrong (e.g. $30$ instead of $0.524$), and no amount of correct method afterwards can recover it. Always check the mode before finding the first solution.
Check yourself
Solve $2\sin^2\theta - 1 = 0$ for $0 \le \theta \le 2\pi$, giving exact answers in terms of $\pi$. (Hint: rearrange to make $\sin\theta$ the subject first — remember this gives two possible values, one positive and one negative.)
Rearrange for sin²θ, then square-root both sides — keeping both signs.
2
For $\sin\theta=\dfrac{1}{\sqrt2}$: principal value $\dfrac{\pi}{4}$, second solution $\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$.
Same sketch-and-reflect method as Example 1, for the positive branch.
3
For $\sin\theta=-\dfrac{1}{\sqrt2}$: by symmetry (sine is negative in the third and fourth quadrants), solutions are $\pi+\dfrac{\pi}{4}=\dfrac{5\pi}{4}$ and $2\pi-\dfrac{\pi}{4}=\dfrac{7\pi}{4}$.
The negative branch mirrors the positive one into the lower half of the sketch.