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Mathematics · Topic 39

Solving trigonometric equations

Sub-topic 1 2025 · National One-Eighth Free

Solving $\cos x=-\dfrac{\sqrt3}{2}$ for $0\lt x\lt\pi$

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Sub-topic 2 2024 · National Preliminary Premium

Solving $2\cos x+\sqrt3=0$ for $0\lt x\lt\pi$

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Sub-topic 3 2026 · National One-Eighth Premium

Solving $2\cos x+\sqrt3=0$ for $\dfrac{\pi}{2}\lt x\lt\pi$

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Sub-topic 4 2025 · National Preliminary Premium

Solving $\sin x=\dfrac{\sqrt2}{2}$ for $0\lt x\lt\pi$

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Sub-topic 5 2025 · National One-Eighth Premium

Solving $2\sin^2x-\sin x=0$ for $0\lt x\lt\pi$

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Sub-topic 6 2025 · National One-Eighth Premium

Solving $2\sin^2x-\sin x=0$ for $0\lt x\lt\pi$

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Sub-topic 7 2026 · National Preliminary Premium

Solving $(2\sin x-1)(\sqrt2\cos x+1)=0$ for $0\lt x\lt\dfrac{\pi}{2}$

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Sub-topic 8 2025 · Regional Championship Premium

Solving factorised sin/cos equations such as $(2\sin x-1)(2\cos x+1)=0$

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Sub-topic 9 2026 · National Quarterfinal Premium

Solving $\sin\left(\theta-\dfrac{\pi}{6}\right)=\dfrac{1}{\sqrt2}$ for $0\lt\theta\lt\pi$

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Sub-topic 10 2026 · National Preliminary Premium

Solving $\sin(2x-60^\circ)=\dfrac{\sqrt3}{2}$ for $x\gt0$

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Sub-topic 11 2026 · National One-Eighth Premium

Solving $2\cos x+\sqrt3=0$ for $\dfrac{\pi}{2}\lt x\lt\pi$

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Sub-topic 12 2026 · National One-Eighth Premium

Solving $(\tan x-1)(\tan x+\sqrt3)=0$ for $0\lt x\lt\pi$

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Sub-topic 13 2026 · National One-Eighth Premium

Solving $\sin x=-\dfrac{1}{2}$ for $\pi\lt x\lt\dfrac{3\pi}{2}$

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Sub-topic 14 2013 · National Semifinal Premium

Solving $\cos^2(x)=\dfrac{1}{4}$ for $x$ in $0^\circ$ to $180^\circ$

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Sub-topic 15 2013 · National Semifinal Premium

Solving $\sin^2(x)=\dfrac{1}{4}$ for $x$ in $0^\circ$ to $180^\circ$

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Sub-topic 16 2013 · National Semifinal Premium

Solving $\tan^2(x)=1$ for $x$ in $0^\circ$ to $180^\circ$

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Sub-topic 17 2021 · National Quarterfinal Premium

Solving $\sin^2x-\cos^2x=1$ for $0\lt x\lt\pi$

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Sub-topic 18 2021 · National Preliminary Premium

Solving $2\sin^2x-\sin x=0$ for $0\lt x\lt\dfrac{\pi}{2}$

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Sub-topic 19 2022 · National Preliminary Premium

Solving $\cos^2x=\dfrac{1}{2}$ for $0\lt x\lt\pi$

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