Find all solutions of the equation in the interval $[0, 2\pi)$. $\tan 3\theta + 1 = \sec 3\theta$

Question

Find all solutions of the equation in the interval $[0, 2\pi)$. Express your answer in terms of $k$, where $k$ is any integer. $$\tan 3\theta + 1 = \sec 3\theta$$

What do I do with the $3\theta$? I manipulated the equation but I don't know what to do now:

$\frac{\sin3\theta}{\cos3\theta} +1=\frac{1}{\cos3\theta}$

yeah but it gave me $sin3\theta+cos3\theta=1$, and I still don't know what to do — TheNewGuy, Aug 06 '15 at 03:17
Continuing on your theme, can you now multiply throughout by $\frac{1}{\sqrt 2}$, then the first one write $\cos \frac{\pi}{4}$ and the second write $\sin \frac{\pi}{4} $, then use the $\sin (A + B)$ formula ? — Shailesh, Aug 06 '15 at 03:24

score 1 · Answer 1 · answered Aug 06 '15 at 03:27

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Multiply through by $\cos3\theta$, then square both sides to obtain

$$(\sin3\theta+\cos3\theta)^2=1\\\sin^23\theta+2\sin3\theta\cos3\theta+\cos^23\theta=1\\ 1+2\sin3\theta\cos3\theta=1\\ 2\sin3\theta\cos3\theta=0\\ \sin6\theta=0$$

Can you take it from here?

answered Aug 06 '15 at 03:27

abiessu

8,115

When squaring, remember to check the solutions you get to ensure that you've not introduced redundant ones. – Deepak Aug 06 '15 at 03:33
$1=-1$ is a false statement but after squaring it becomes a true statement $1=1$ – Tucker Aug 06 '15 at 03:36

score 1 · Accepted Answer · answered Aug 06 '15 at 03:37

1

Divide throughout by $\sec 3\theta$ to give $\sin 3\theta + \cos 3\theta = 1$.

Now observe that the LHS is basically $\sqrt 2 \sin (3 \theta + \frac{\pi}{4})$

So,

$\sin (3 \theta + \frac{\pi}{4}) = \frac{1}{\sqrt 2}$

$3 \theta + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi, k \in \mathbb{Z}$

Hence $\theta = \frac{2k\pi}{3}, k \in \mathbb{Z}$

In the required range, you have the solutions $\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$.

answered Aug 06 '15 at 03:37

Deepak

26,801

Thank you, that makes sense! – TheNewGuy Aug 06 '15 at 04:09
@TheNewGuy, See also, http://math.stackexchange.com/questions/618192/how-to-solve-sin-x-cos-x-1 – lab bhattacharjee Aug 06 '15 at 05:02
@TheNewGuy There's one more detail I forgot. You need to justify the division by stating that $\sec 3\theta \neq 0$ (the secant can never equal zero for any argument). – Deepak Aug 06 '15 at 05:03

score 0 · Answer 3 · answered Aug 06 '15 at 03:22

0

Hint:

You can square both sides and use the identity,

$$1 + \tan^2 \phi = \sec^2 \phi $$

which simplifies things a lot. But make sure to recheck whatever solution you obtain from the squared equation by substitution in the original one.

answered Aug 06 '15 at 03:22

Ishfaaq

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That downvote's not cool, no? – Ishfaaq Aug 06 '15 at 05:55

score 0 · Answer 4 · answered Aug 06 '15 at 04:38

$$\sec2A-\tan2A=\dfrac{1-\sin2A}{\cos2A}=\dfrac{(\cos A-\sin A)^2}{\cos^2A-\sin^2A}$$

If $\cos A-\sin A=0\iff\tan A=1\implies\sec2A,\tan2A$ are $\infty$

So, $\cos A-\sin A\ne0$

$$\sec2A-\tan2A=\dfrac{\cos A-\sin A}{\cos A+\sin A}=\dfrac{1-\tan A}{1+\tan A}=\tan\left(\dfrac\pi4-A\right)$$

$$\tan\left(\dfrac\pi4-A\right)=1=\tan\dfrac\pi4\implies\dfrac\pi4-A=n\pi+\dfrac\pi4$$ where $n$ is any integer

Here $2A=3\theta$

Find all solutions of the equation in the interval $[0, 2\pi)$. $\tan 3\theta + 1 = \sec 3\theta$

4 Answers4