How to solve this Trigonometric equation $\tan^{2}\theta + \sec(2\theta)=1$?
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.
3 Answers
$\begingroup$$$\tan^2t=1-\sec2t=1-\dfrac{1+\tan^2t}{1-\tan^2t}=-\dfrac{2\tan^2t}{1-\tan^2t}$$
$$\iff\tan^2t(3-\tan^2t)=0$$
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
$\endgroup$ $\begingroup$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).
$\endgroup$ $\begingroup$$$\tan^{2}\theta + \sec2\theta=1$$ $$\frac {\sin^{2}\theta}{\cos^{2}\theta}+\frac1{\cos2\theta}=1$$ $$\frac{1-\cos2\theta}{1+cos2\theta}+\frac1{\cos2\theta}=1$$ $\cos2\theta=1$ 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 $\theta=\pm \frac{\pi}3 +\pi n, m,n\in \mathbb Z$
$\endgroup$ 0
