Every solution from 0° to 360°
An equation like sin x = k \sin x = k sin x = k 0 ∘ 0^\circ 0 ∘ 360 ∘ 360^\circ 36 0 ∘ reference angle first, then use the quadrants where the ratio has the right sign:
All Sin Tan Cos 0° 90° 180° 270° Where each ratio is positive All, Sin, Tan, Cos: going anticlockwise from 0°
Second quadrant: 180 ∘ − α 180^\circ - \alpha 18 0 ∘ − α 180 ∘ + α 180^\circ + \alpha 18 0 ∘ + α 360 ∘ − α 360^\circ - \alpha 36 0 ∘ − α
A ratio of 0 0 0 1 1 1 − 1 -1 − 1 0 ∘ 0^\circ 0 ∘ 90 ∘ 90^\circ 9 0 ∘ 180 ∘ 180^\circ 18 0 ∘ 270 ∘ 270^\circ 27 0 ∘ 360 ∘ 360^\circ 36 0 ∘
sin x \sin x sin x cos x \cos x cos x − 1 -1 − 1
Solving a trig equation Pick the ratio and set k
90 180 270 360 −1 −0.5 0.5 1 x° y 36.87°, 143.13° solutions 36.87° reference angle
sin x = 0.6. The reference angle is 36.87°. Sine is positive in the first and second quadrants: ref and 180° − ref. So x = 36.87° or 143.13°. More: every solution from 0° to 360°
Quadratic equations in sin or cos
Treat sin x \sin x sin x cos x \cos x cos x − 1 -1 − 1 1 1 1
Worked example · WAEC 2022
WAEC 2022 · Paper 2 · Q4
Solve, correct to the nearest degree, 3 cos 2 θ + 10 cos θ − 8 = 0 3\cos^2\theta + 10\cos\theta - 8 = 0 3 cos 2 θ + 10 cos θ − 8 = 0 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Factorise
( 3 cos θ − 2 ) ( cos θ + 4 ) = 0 {(3\cos\theta - 2)(\cos\theta + 4) = 0} ( 3 cos θ − 2 ) ( cos θ + 4 ) = 0 cos θ = − 4 {\cos\theta = -4} cos θ = − 4 cos θ = 2 3 {\cos\theta = \frac23} cos θ = 3 2 Think first. Let c = cos θ: 3c² + 10c − 8 = 0.
Every solution
θ = cos − 1 2 3 ≈ 48.19 ∘ {\theta = \cos^{-1}\frac23 \approx 48.19^\circ} θ = cos − 1 3 2 ≈ 48.1 9 ∘ 360 ∘ − 48.19 ∘ = 311.81 ∘ {360^\circ - 48.19^\circ = 311.81^\circ} 36 0 ∘ − 48.1 9 ∘ = 311.8 1 ∘ To the nearest degree: 48 ∘ {48^\circ} 4 8 ∘ 312 ∘ {312^\circ} 31 2 ∘
Think first. Cosine is positive in the first and fourth quadrants.
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When an equation mixes cos 2 x \cos 2x cos 2 x sin x \sin x sin x cos 2 x \cos 2x cos 2 x 1 − 2 sin 2 x 1 - 2\sin^2 x 1 − 2 sin 2 x cos 2 x \cos 2x cos 2 x cos x \cos x cos x 2 cos 2 x − 1 2\cos^2 x - 1 2 cos 2 x − 1
Worked example · WAEC 2011
WAEC 2011 · Paper 2 · Q1
Find the truth set of sin θ + cos 2 θ = 0 \sin\theta + \cos2\theta = 0 sin θ + cos 2 θ = 0 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
One ratio
sin θ + 1 − 2 sin 2 θ = 0 {\sin\theta + 1 - 2\sin^2\theta = 0} sin θ + 1 − 2 sin 2 θ = 0 So 2 sin 2 θ − sin θ − 1 = 0 {2\sin^2\theta - \sin\theta - 1 = 0} 2 sin 2 θ − sin θ − 1 = 0
Think first. The equation has sin θ, so use cos 2θ = 1 − 2 sin²θ.
Factorise
( 2 sin θ + 1 ) ( sin θ − 1 ) = 0 {(2\sin\theta + 1)(\sin\theta - 1) = 0} ( 2 sin θ + 1 ) ( sin θ − 1 ) = 0 sin θ = 1 {\sin\theta = 1} sin θ = 1 sin θ = − 1 2 {\sin\theta = -\frac12} sin θ = − 2 1 Every solution
sin θ = 1 {\sin\theta = 1} sin θ = 1 θ = 90 ∘ {\theta = 90^\circ} θ = 9 0 ∘ sin θ = − 1 2 {\sin\theta = -\frac12} sin θ = − 2 1 θ = 210 ∘ {\theta = 210^\circ} θ = 21 0 ∘ 330 ∘ {330^\circ} 33 0 ∘ Think first. sin θ = −1/2: reference angle 30°, third and fourth quadrants.
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Worked example · WAEC 2022
WAEC 2022 · Paper 2 · Q4
Solve 3 cos 2 x − sin x = 0 3\cos2x - \sin x = 0 3 cos 2 x − sin x = 0 0 ∘ ≤ x ≤ 360 ∘ 0^\circ \le x \le 360^\circ 0 ∘ ≤ x ≤ 36 0 ∘
A quadratic in sin x
3 ( 1 − 2 sin 2 x ) − sin x = 0 {3(1 - 2\sin^2 x) - \sin x = 0} 3 ( 1 − 2 sin 2 x ) − sin x = 0 6 sin 2 x + sin x − 3 = 0 {6\sin^2 x + \sin x - 3 = 0} 6 sin 2 x + sin x − 3 = 0 The formula
sin x = − 1 ± 1 + 72 12 = − 1 ± 73 12 {\sin x = \frac{-1 \pm \sqrt{1 + 72}}{12} = \frac{-1 \pm \sqrt{73}}{12}} sin x = 12 − 1 ± 1 + 72 = 12 − 1 ± 73 sin x ≈ 0.6287 {\sin x \approx 0.6287} sin x ≈ 0.6287 sin x ≈ − 0.7953 {\sin x \approx -0.7953} sin x ≈ − 0.7953 Think first. It doesn't factorise: use the quadratic formula.
Four solutions
sin x = 0.6287 {\sin x = 0.6287} sin x = 0.6287 x ≈ 38.95 ∘ {x \approx 38.95^\circ} x ≈ 38.9 5 ∘ 180 ∘ − 38.95 ∘ = 141.05 ∘ {180^\circ - 38.95^\circ = 141.05^\circ} 18 0 ∘ − 38.9 5 ∘ = 141.0 5 ∘ sin x = − 0.7953 {\sin x = -0.7953} sin x = − 0.7953 52.69 ∘ {52.69^\circ} 52.6 9 ∘ x ≈ 232.69 ∘ {x \approx 232.69^\circ} x ≈ 232.6 9 ∘ 307.31 ∘ {307.31^\circ} 307.3 1 ∘ ← Back Next step → Show all steps Start again
More: trigonometric equations
Your turn
(b) Solve the equation 2 cos 2 θ − 5 cos θ = 3 2\cos^2\theta - 5\cos\theta = 3 2 cos 2 θ − 5 cos θ = 3 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Worked solution (try it first) (b) Let
c = cos θ c = \cos\theta c = cos θ :
2 c 2 − 5 c − 3 = 0 2c^2 - 5c - 3 = 0 2 c 2 − 5 c − 3 = 0 , so
( 2 c + 1 ) ( c − 3 ) = 0 (2c + 1)(c - 3) = 0 ( 2 c + 1 ) ( c − 3 ) = 0 .
cos θ = 3 \cos\theta = 3 cos θ = 3 is impossible, so
cos θ = − 1 2 \cos\theta = -\frac12 cos θ = − 2 1 .
Cosine is negative in the second and third quadrants:
θ = 180 ∘ − 60 ∘ = 120 ∘ \theta = 180^\circ - 60^\circ = 120^\circ θ = 18 0 ∘ − 6 0 ∘ = 12 0 ∘ or
θ = 180 ∘ + 60 ∘ = 240 ∘ \theta = 180^\circ + 60^\circ = 240^\circ θ = 18 0 ∘ + 6 0 ∘ = 24 0 ∘ .
Watch out
In (a), reject k = − 8 k = -8 k = − 8 k k k In (b), cos θ = − 1 2 \cos\theta = -\frac12 cos θ = − 2 1 0 ∘ 0^\circ 0 ∘ 360 ∘ 360^\circ 36 0 ∘ 120 ∘ 120^\circ 12 0 ∘ Report a problem with this question