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I'm learning trigonometry for a maths exam in college and I'm just wondering if someone could give a step by step tutorial on how to solve a typical trigonometry equation like this or send a link for a decent tutorial on trigonometric equations. I've looked for tutorials else where but haven't come across any good ones.

(a)Solve $3\tan𝜃 + 5 = 7$ for $0 ≤ 𝜃 ≤ 360$

(b)Solve $\cos3𝜃 = -0.5$ for $0 ≤ 𝜃 ≤ 360$

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  • $\begingroup$ Welcome to Math.SE! What background do you come from? Have you taken classes on this subject? Do you own any materials to provide context for prior knowledge? $\endgroup$ – Brevan Ellefsen May 1 '17 at 20:23
  • $\begingroup$ @BrevanEllefsen Thank you, I'm doing a course in Software Engineering. I have very basic trigonometry skills from high school as I unfortunately missed these lectures through illness. $\endgroup$ – Will Mannix May 1 '17 at 20:37
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Think of $\tan\theta$ or $\cos 3\theta$ as a variable instead of a function. It may even help to actually replace it with a more familiar looking variable. i.e. $u = \tan \theta$

Isolate it.

Once you have isolated it, you can use apply the inverse function.

e.g.

$3\tan\theta + 5 = 7\\ u = \tan\theta\\ 3u + 5 = 7\\ u = \frac {2}{3}\\ \tan\theta = \frac {2}{3}\\ \theta = \arctan (\frac {2}{3})$

This is not one of your special angles, so you will need a calculator. $\theta = 33.69^\circ$

$\tan \theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\circ)$. There is second answer in the interval $[0,360^\circ)$.

$\tan \theta = \tan (\theta + 180)$

$213.69^\circ$ is also in the solution set.

The next one might even be a little bit trickier.

$\cos 3\theta = -0.5\\ 3\theta = \arccos (-0.5)\\ 3\theta = 120^\circ$

but again, this isn't the only answer.

$\theta = \arccos (-0.5)$ has two answers in the 360 degree interval, and $3\theta = \arccos (-0.5)$ will have $6.$

$\cos \theta = \cos (-\theta)$ and $\cos \theta = \cos (360+\theta)$ using these identities we find other values of theta.

$3\theta = 240^\circ, 480^\circ, 600^\circ, 840^\circ, 960^\circ $ are all viable, as well.

$\theta \in \{40, 80, 160, 200, 280, 320\}$

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a) Reduce to $$\tan\theta=\frac23.$$

Then as the tangent function has period $180°$ and is monotonic in a period,

$$\tan\theta=\tan(\theta+k180°)$$ and all solutions are of the form

$$\theta+k180°=\arctan\frac23.$$

Find all $k$ such that $0\le\theta\le360°$.

b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that

$$\cos3\theta=\cos(k360°\pm3\theta).$$

Then all solutions are of the form

$$k120°\pm\theta=\frac13\arccos(-0.5).$$

Again, find all $k$ such that $0\le\theta\le360°$.


It helps a lot to draw a trigonometric circle to discuss the solutions.

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