$$\tan\theta + \sec\theta =2\cos \theta,\quad 0\le \theta\le 2\pi$$Find all the possible solutions for the equations.

Multiply both sides by $\sec\theta - \tan \theta$. $$\implies (\tan\theta + \sec\theta)(\sec\theta - \tan\theta) = (\sec\theta -\tan\theta)2\cos \theta$$ $$\implies 1 = 2 -2\sin \theta$$ $$\implies \sin \theta=\frac12 \implies \theta = \arcsin\frac12$$Such a solution gets me two solutions $\frac{\pi}6$ and $\frac{5\pi}6$. But when I Wolfram it, I am supposed to get one more solution i.e $\frac{3\pi}2$, but at $\frac{3\pi}2$ $\tan \theta$ and $\sec\theta$ aren't defined.

  • $\begingroup$ What exactly did you "Wolfram"? And, of course, $3\pi/2$ is not a solution of your equation. $\endgroup$ – David Mitra Apr 30 '12 at 19:20
  • $\begingroup$ I think Wolfram simplifies by multiplying by $\cos(\theta)$... $\endgroup$ – draks ... Apr 30 '12 at 19:22
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – MSC Apr 30 '12 at 19:23
  • 1
    $\begingroup$ -1 for Wolfram... $\endgroup$ – David Mitra Apr 30 '12 at 19:31
  • 4
    $\begingroup$ @ManiSarkarCallisto: I would not worry unduly about what Wolfram Alpha thinks. $\endgroup$ – André Nicolas Apr 30 '12 at 19:55

Start with $$\tan(\theta) + \sec(\theta) = 2\cos(\theta).$$ Multiply both sides by $\cos(\theta)$ to get $$\sin(\theta) + 1 = 2\cos^2(\theta);$$ be warned that extraneous roots could be introduced where $\cos(\theta) = 0$, so you will hve to check these two roots separately.

Now use the Pythagorean identity to get

$$\sin(\theta) + 1 = 2 - 2\sin^2(\theta).$$ This is a quadratic-in-drag. Solve it; then check the two other places where $\cos(\theta) = 0$ separately. Beware of any domain considerations.



$$\implies \sec\theta-\tan\theta=\frac1{\sec\theta+\tan\theta}=\frac{\sec\theta}2$$

$$\implies \sec\theta\left(1-\frac12\right)=\tan\theta$$

$$\implies \frac{\tan\theta}{\sec\theta}=\frac12\iff \sin\theta=\frac12=\sin\frac\pi6$$

$$\implies \theta=n\pi+(-1)^n\frac\pi6\text{ where }n \text{ is any integer}$$

If $n$ is even $=2m$(say), $\theta=2m\pi+\frac\pi6$

As $0\le \theta\le2\pi, 0\le2m\pi+\frac\pi6\le2\pi\implies 0\le 12m+1\le12\implies m=0$

Similarly, if $n$ is odd $=2m+1$(say), $\theta=(2m+1)\pi-\frac\pi6\implies m=0$

  • $\begingroup$ @MSC, try to avoid division & multiplication or squaring which often introduce extraneous roots which needs elimination $\endgroup$ – lab bhattacharjee May 12 '13 at 7:03

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