Simple Trigonometric Equations

What this move does

There are only three basic trigonometric equations, and the unit circle shows their solutions at a glance: at most two points in one turn, plus the period:

sin x = a cos x = a tan x = a \Rightarrow x = k π + (- 1)^{k} arcsin a \Rightarrow x = 2 k π \pm arccos a \Rightarrow x = k π + arctan a

Here $k \in Z$ ; the first two need $∣ a ∣ \leq 1$ , otherwise there is no solution.

Use factoring, substitution, the auxiliary angle and so on to reduce the equation to one or more basic equations.
For each basic equation, find the solutions in one turn of the unit circle (discard any with $∣ a ∣ > 1$ straight away).
Add the period to write the general solution; if the problem gives an interval, count the solutions in that interval carefully.

Predict first

How many solutions does $sin 3 x = \frac{1}{2}$ have in $[0, 2 π)$ ?

Choose an answer first; then you can see the result.

Work through one

Example: solve 2sin²x − sin x − 1 = 0, x ∈ [0, 2π)

Step 1 / 5

Substitute
Treat $sin x$ as a single unknown $t$ , and the equation becomes a quadratic. Note that $t$ can only take values in $[- 1, 1]$ .

$2 t^{2} - t - 1 = 0 ⟹ (t - 1) (2 t + 1) = 0$
The first basic equation
The second basic equation
Put them together
If some t falls outside [−1, 1]

Cosine and tangent work the same way; only the direction of the line cutting the circle changes. For $cos x = a$ use the vertical line $x = a$ ; for $tan x = a$ use the line through the origin with slope $a$ (it meets the circle at two points that differ by $π$ , which is why the period of tangent is $π$ ).

Switch to cos or tan and drag

a

. The tangent line always meets the circle at two diametrically opposite points.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: $sin ω x = a$ has at most $2$ solutions in $[0, 2 π)$

When $ω = 1$ , as $x$ runs through $[0, 2 π)$ it goes around the unit circle exactly once, so there are never more than two solutions.

Practice

Exercise 1Basic

Solve $cos x = - \frac{2}{2}$ in $[0^{\circ}, 36 0^{\circ})$ (answer in degrees, separated by commas).

Exercise 2Basic

Solve $2 sin^{2} x + 3 sin x - 2 = 0$ in $[0^{\circ}, 36 0^{\circ})$ .

Exercise 3Advanced

Solve $sin 2 x = cos x$ in $[0^{\circ}, 36 0^{\circ})$ .

Exercise 4Challenge

Solve $sin x + 3 cos x = 1$ in $[0^{\circ}, 36 0^{\circ})$ .

What this move does

Work through one

Example: solve 2sin²x − sin x − 1 = 0, x ∈ [0, 2π)

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Dividing both sides by sin x and losing roots