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NSWMaths Extension 1Quick questions

Trigonometric Functions (ME-T1, T2, T3)

Quick questions on General solutions of trigonometric equations: sin⁡\sin, cos⁡\cos and tan⁡\tan

14short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is general solution for sin⁡θ=k\sin \theta = k (with −1≤k≤1-1 \le k \le 1)?
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sin⁡\sin has period 2π2 \pi and is symmetric about θ=π2\theta = \frac{\pi}{2}: sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin \theta. So solutions are
What is general solution for cos⁡θ=k\cos \theta = k (with −1≤k≤1-1 \le k \le 1)?
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cos⁡\cos has period 2π2 \pi and is even: cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta. So solutions are
What is general solution for tan⁡θ=k\tan \theta = k?
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tan⁡\tan has period π\pi (not 2π2 \pi). Solutions are
What is equations with composite arguments?
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For sin⁡(2x+1)=12\sin(2 x + 1) = \frac{1}{2}, treat u=2x+1u = 2 x + 1 as the variable, find the general solution for uu, then back-solve for xx.
What is restriction to an interval?
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After writing the general solution, list the values of nn that put θ\theta in the required interval. Use a number line or trial substitution.
What is multiple-step equations?
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Combine general-solution skills with identities:
What is cosine equation?
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Find all xx in [0,2π][0, 2 \pi] with cos⁡x=−12\cos x = -\frac{1}{2}.
What is tan equation?
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Solve tan⁡x=1\tan x = 1 for x∈[0,2π]x \in [0, 2 \pi].
What is composite argument?
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Solve cos⁡ ⁣(2x+π3)=22\cos\!\left( 2 x + \frac{\pi}{3} \right) = \frac{\sqrt{2}}{2} for x∈[0,π]x \in [0, \pi].
What is factorable equation?
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Solve 2sin⁡xcos⁡x=sin⁡x2 \sin x \cos x = \sin x for x∈[0,2π]x \in [0, 2 \pi].
What is treating tan⁡\tan like sin⁡\sin or cos⁡\cos?
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tan⁡\tan has period π\pi, not 2π2 \pi. The general solution is α+nπ\alpha + n \pi, not ±α+2nπ\pm \alpha + 2 n \pi.
What is dividing by zero?
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Solving sin⁡xcos⁡x=sin⁡x\sin x \cos x = \sin x by dividing by sin⁡x\sin x loses the solutions where sin⁡x=0\sin x = 0. Always factor instead.
What is forgetting to back-out the substitution?
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For sin⁡(2x)=k\sin(2 x) = k, you get 2x=2 x = general expression, then divide by 22 at the end. The division spans every term in the general expression, including 2nπ2 n \pi.
What is off-by-one on the interval?
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Endpoints matter: [0,2π)[0, 2 \pi) excludes 2π2 \pi while [0,2π][0, 2 \pi] includes it. Check the question wording. :::

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