Solving trig equations — general solutions, solutions in given range: Question bank

True or false — justify

cos θ = \frac{1}{2}

has exactly two solutions in

0 \leq θ < 36 0^{\circ}

True — cosine hits each value (except

\pm 1

) exactly twice per turn, here at

6 0^{\circ}

and

30 0^{\circ}

, the pair mirrored across the

x

-axis.

sin θ = 2

has infinitely many solutions because sine is periodic.

False — periodicity only multiplies existing solutions. Sine never exceeds

1

, so there are zero solutions, and repeating nothing still gives nothing.

The general solution

θ = 2 nπ \pm α

and

θ = - 2 nπ \pm α

describe the same set.

True — as

n

runs over all integers (

0, \pm 1, \pm 2, \dots

2 nπ

and

- 2 nπ

sweep the same multiples of

2 π

, so both formulas produce the identical family.

For

sin θ = sin α

, the two per-turn solutions are

α

and

- α

False — they are

α

and

π - α

(supplementary angles), not

\pm α

. The

\pm α

pattern belongs to cosine, which is even; sine is symmetric about

θ = \frac{π}{2}

instead.

tan θ = 5

has only one solution in

0 \leq θ < 2 π

False — tan has period

π

, so a value repeats every half-turn;

tan θ = 5

has two solutions in a full

2 π

(one in the first quadrant, one in the third).

Changing an equation from degrees to radians changes the number of solutions in "one full turn".

False — one full turn is

36 0^{\circ} = 2 π

radians; it's the same physical range, so the count is identical. Only the numbers labelling the answers change.

cos θ = - 1

has two solutions in each

2 π

interval.

False —

- 1

is the minimum of cosine, touched at exactly one point per turn (

18 0^{\circ}

). The "two per turn" rule fails at the peak and trough where the pair collapses into one.

Spot the error

cos θ = \frac{1}{2} \Rightarrow θ = 6 0^{\circ}

, done."

The calculator returns only the principal value. The full family is

θ = 36 0^{\circ} n \pm 6 0^{\circ}

; filtering to

0 \leq θ < 36 0^{\circ}

also gives

30 0^{\circ}

, which was silently dropped.

sin θ = \frac{3}{2}

, so

θ = nπ \pm \frac{π}{3}

Sine uses

nπ + (- 1)^{n} α

, not

\pm α

. The

\pm

form would wrongly hand you

- \frac{π}{3}

-type angles where sine is negative, so it generates false roots.

cos 2 θ = \frac{3}{2}

0 \leq θ < 2 π

, so I solve

cos ϕ = \frac{3}{2}

0 \leq ϕ < 2 π

with

ϕ = 2 θ

Substituting

ϕ = 2 θ

doubles the interval to

0 \leq ϕ < 4 π

. Solving only in

[0, 2 π)

keeps half the

ϕ

values, so you lose half the final

θ

answers.

tan θ = - 1

, calculator gives

- 4 5^{\circ}

, so one answer is

- 4 5^{\circ}

and I stop."

- 4 5^{\circ}

is outside

0^{\circ} \leq θ < 36 0^{\circ}

. You must add multiples of the period

18 0^{\circ}

to land inside:

13 5^{\circ}

and

31 5^{\circ}

are the actual in-range solutions.

"For

cos 2 θ = \frac{3}{2}

, I found

ϕ

, and

θ = ϕ + 2 nπ

after dividing."

Divide every solved

ϕ

2

directly (

θ = ϕ /2

); don't re-add the period after dividing. The extra turns were already accounted for when you widened the

ϕ

-range.

sin θ = 0

gives

θ = nπ + (- 1)^{n} \cdot 0 = nπ

, and each of these is a distinct pair."

The formula is fine and gives

θ = nπ

, but since

α = 0

the "two families" merge —

α

and

π - α

coincide at

0

and

π

, so there's just one solution per

π

, not two per

2 π

2 sin θ cos θ = 1

has no nice solution, so I'll square both sides."

Squaring introduces extra roots (it also solves

2 sin θ cos θ = - 1

). Instead recognise

2 sin θ cos θ = sin 2 θ

via Compound and Double Angle Formulae and solve directly.

Why questions

Why does cosine use

\pm α

while sine uses

(- 1)^{n} α

Cosine is even (

cos (- θ) = cos θ

), so its two solutions mirror across the

x

-axis as

\pm α

. Sine's two solutions are supplementary (

α

and

π - α

); the

(- 1)^{n}

toggle packs that alternation into one formula.

Why does tangent need only one family (no

\pm

, no alternation)?

Because

tan

has period

π

and, within each period, is strictly increasing — it hits every value exactly once. One base angle

α

plus

nπ

therefore captures everything, no second family required.

Why must we "widen the range" before dividing in multiple-angle problems?

The unknown inside is

ϕ = k θ

; as

θ

crosses its interval,

ϕ

crosses a

k

-times-longer interval. Solving

ϕ

over the un-widened range hides the extra periods that map to valid

θ

values.

Why can't we just add

36 0^{\circ}

once and assume we've found all solutions?

One period-shift catches the next copy, but the range may span several periods (especially after widening for multiple angles). You must test

n = \dots, - 1, 0, 1, 2, \dots

until the outputs leave the range on both ends.

Why does

cos θ = c

have no solutions when

∣ c ∣ > 1

On the unit circle,

cos θ

is the horizontal coordinate of a point on a circle of radius

1

, so it can never exceed

1

in size. No point sits at horizontal distance

> 1

Why is the "principal value" from an inverse function unique even though the equation isn't?

Inverse trig functions are defined on a restricted domain (e.g.

cos^{- 1}

[0, π]

) precisely so they return one answer. The general solution then rebuilds the full family from that single anchor

α

Why do we look at the graph to predict the number of solutions before solving?

Drawing

y = sin θ

and the horizontal line

y = c

over the range, each crossing is one solution — see Graphs of Trigonometric Functions. This sanity-checks that your general-solution filtering kept the right count.

Edge cases

cos θ = 1

— how many solutions in

0 \leq θ < 2 π

, and why is it not two?

Exactly one (

θ = 0

). At the maximum the two mirrored solutions

\pm α

collapse into the single point

α = 0

, so the pair degenerates.

sin θ = 1

— what happens to the two supplementary solutions?

They merge:

α = \frac{π}{2}

and

π - α = \frac{π}{2}

are the same angle. So there is one solution per turn, at the peak of the sine curve.

tan θ = 0

— how does this differ from

tan θ = t

for a generic

t

It doesn't degenerate:

θ = nπ

still gives one solution per period. But note

tan θ = 0

exactly where

sin θ = 0

, and tan is undefined (not zero) at

θ = \frac{π}{2} + nπ

cos θ = 0

— is

θ = \frac{π}{2}

the only solution per turn?

No — cosine crosses zero twice per turn, at

\frac{π}{2}

and

\frac{3 π}{2}

. Zero is a mid-range value, not a peak, so the pair does not collapse.

A range like

0^{\circ} \leq θ \leq 36 0^{\circ}

(closed at both ends) versus

0^{\circ} \leq θ < 36 0^{\circ}

— does it matter?

Yes at the boundary. If

θ = 0^{\circ}

solves it, then

36 0^{\circ}

also solves it; the closed range would list both, the half-open range only

0^{\circ}

. Always check whether endpoints are included.

tan θ = t

where the required range straddles an asymptote (e.g. near

9 0^{\circ}

) — any trap?

The asymptote itself is not a solution (tan is undefined there), but solutions exist just below and just above it. Don't skip the interval around

9 0^{\circ}

; just exclude the exact asymptote value.

What if an equation reduces to something like

(sin θ - 1) (2 cos θ - 1) = 0

Solve each factor separately as its own equation (

sin θ = 1

and

cos θ = \frac{1}{2}

), then combine all in-range answers. A product is zero when any factor is zero, so missing a factor loses solutions — use Trigonometric Identities to reach such factored forms.

Connections

Unit Circle and Radian Measure — the geometric source of $\pm$ , supplements, and the $∣ c ∣ \leq 1$ bound.
Graphs of Trigonometric Functions — count solutions by counting line-crossings.
Inverse Trigonometric Functions — why the principal value $α$ is unique.
Trigonometric Identities — factor and simplify before solving.
Compound and Double Angle Formulae — turn multiple-angle traps into solvable forms.