IP AMaths Notes (Upper Sec, Year 3-4): 13) Trigonometry II

With radian fluency in hand, solve trig equations by mapping solutions across quadrants and applying compound-angle identities.

Identities to deploy

• Compound: \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \).
• Compound: \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \).
• Auxiliary angle: express \( a \sin x + b \cos x = R \sin(x + \alpha) \) where \( R = \sqrt{a^2 + b^2} \) and \( \alpha = \arctan\bigl(\dfrac{b}{a}\bigr) \).

Worked example 1 — Solve trig equation

Solve \( 2 \sin^2 x - 3 \sin x + 1 = 0 \) for \( 0 \leq x < 2\pi \).

1. Let \( y = \sin x \). Equation becomes \( 2y^2 - 3y + 1 = 0 \).
2. Factor: \( (2y - 1)(y - 1) = 0 \).
3. So \( y = \dfrac{1}{2} \) or \( y = 1 \).
4. If \( \sin x = \dfrac{1}{2} \), then \( x = \dfrac{\pi}{6} \) or \( x = \dfrac{5\pi}{6} \).
5. If \( \sin x = 1 \), then \( x = \dfrac{\pi}{2} \).

Solutions: \( x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6} \).

Worked example 2 — Auxiliary angle

Solve \( 3 \sin x + 4 \cos x = 2 \) for \( 0 \leq x < 2\pi \).

1. Compute \( R = \sqrt{3^2 + 4^2} = 5 \).
2. Let \( \alpha = \arctan\bigl(\tfrac{4}{3}\bigr) \).
3. Rewrite: \( 3 \sin x + 4 \cos x = 5 \sin(x + \alpha) \).
4. Equation becomes \( \sin(x + \alpha) = \tfrac{2}{5} \).
5. Principal solution: \( x + \alpha = \arcsin\bigl(\tfrac{2}{5}\bigr) \).
6. General solutions within range: \( x + \alpha = \arcsin\bigl(\tfrac{2}{5}\bigr) \) or \( x + \alpha = \pi - \arcsin\bigl(\tfrac{2}{5}\bigr) \).
7. Subtract \( \alpha \) to get \( x \). Numerically, \( \alpha = 0.927\) rad, \( \arcsin\bigl(\tfrac{2}{5}\bigr) = 0.411\) rad.
8. Hence \( x \approx -0.516 \) rad; If we add \( 2\pi \) → \( 5.767 \) rad. Or take the second branch \( x \approx 1.803 \) rad.

Try this

Solve \( 2 \cos 2x = \sqrt{3} \) for \( 0 \leq x < 2\pi \), displaying answers in exact form.