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What is the general solution of this trigonometric equation $$\tan^{2}\theta + \sec(2\theta)=1$$ from the following options:

a) $m\pi$

b) $n\pi\pm \frac{\pi}{3}$

c) $m\pi,n\pi\pm \frac{\pi}{3},\text{where }m,n\in \mathbb{Z}$

d) None of these

By examining the options in the given equation I get option a) b) c) satisfy the equation but I can't simplify it to get a general solution by myself.
My attempt:
$$\sec^{2}\theta -1+ \sec(2\theta)=1$$ Changing secant in terms of cosecant also doesn't help much instead make it complicated. I couldn't solve it further, thanks for help.

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    $\begingroup$ You have to have some exposure to manipulating trig functions. Just play with it for at least five minutes to see how you can change it before giving up. Hint: Try the double angle formula? $\endgroup$ – MPW May 13 '16 at 7:38
  • $\begingroup$ Hint: Looks like a double angle formula might be useful. $\endgroup$ – Karl May 13 '16 at 7:39
  • $\begingroup$ what means $m,n\in I$? $\endgroup$ – miracle173 May 13 '16 at 9:15
  • $\begingroup$ Thanks for asking, used the wrong notation, correcting it. I need to put integers there $\endgroup$ – Onix May 13 '16 at 9:55
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$$\tan^{2}\theta + \sec2\theta=1$$ $$\frac {\sin^{2}\theta}{\cos^{2}\theta}+\frac1{\cos2\theta}=1$$ $$\frac{1-\cos2\theta}{1+cos2\theta}+\frac1{\cos2\theta}=1$$ $\cos2\theta=1$ or $\cos2\theta=-\frac12$

$2\theta=2\pi m$ or $2\theta=\pm \frac{2\pi}3 +2\pi n, m,n\in \mathbb Z$

$\theta=\pi m$ or $\theta=\pm \frac{\pi}3 +\pi n, m,n\in \mathbb Z$

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$$\tan^2t=1-\sec2t=1-\dfrac{1+\tan^2t}{1-\tan^2t}=-\dfrac{2\tan^2t}{1-\tan^2t}$$

$$\iff\tan^2t(3-\tan^2t)=0$$

Either $$\tan^2t=0\iff\tan t=0\iff t=m\pi$$

OR $$\tan^2t=3=\tan^2\dfrac\pi3$$

We know if $$\tan^2y=\tan^2A\iff y=r\pi\pm A$$ where $m,r$ are integers

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Use $\cos2\theta=2\cos^2\theta-1$ (and of course $\sin^2\theta=1-\cos^2\theta$). Putting $c=\cos\theta$, we have $\frac{1-c^2}{c}+\frac{1}{2c^2-1}=1$. Simplifying this becomes $(4c^2-1)(c^2-1)=0$, so $c=\pm1,\pm\frac{1}{2}$, giving the roots in (c).

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