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$$\tan\theta + \sec\theta =2\cos \theta,\quad 0\le \theta\le 2\pi$$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.

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    What exactly did you "Wolfram"? And, of course, $3\pi/2$ is not a solution of your equation.2012-04-30
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    I think Wolfram simplifies by multiplying by $\cos(\theta)$...2012-04-30
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    http://www.wolframalpha.com/input/?i=tan%28%CE%B8%29+%2B+sec%28%CE%B8%29+%3D2cos%28%CE%B8%29%2C+0%E2%89%A4%CE%B8%E2%89%A42%CF%802012-04-30
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    -1 for Wolfram...2012-04-30
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    @DavidMitra and draks, Can you please explain me why Wolfram is wrong?2012-04-30
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    Multiplying both sides of the original equation by $\cos \theta$ gives $\sin\theta+1=2\cos^2\theta$. This has $\theta=3\pi/2$ as a solution.2012-04-30
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    @DavidMitra What about this? How could Wolfram even plot a function wrong? http://www.wolframalpha.com/input/?i=Plot+tan%CE%B8%2Bsec%CE%B8-2cos%CE%B8%2C+0%E2%89%A4%CE%B8%E2%89%A42%CF%802012-04-30
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    The limit as $\theta\rightarrow 3\pi/2$ of $\tan\theta-\sec\theta$ is 0 (in fact Wolfram returns $0$ for the expression $\tan(3pi/2)-\sec( 3pi/2)$). I'm not sure why Wolfram is regarding $3\pi/2$ as a valid solution of your equation. As you said, neither $\tan$ nor $\sec$ is defined at $3\pi/2$.2012-04-30
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    @DavidMitra Surprisingly, $\sec\frac{3\pi}2 - \tan\frac{3\pi}2$ is defined. Is this really possible? http://www.wolframalpha.com/input/?i=sec%283pi%2F2%29+-+tan%283pi%2F2%292012-04-30
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    @ManiSarkarCallisto: I would not worry unduly about what Wolfram Alpha thinks.2012-04-30

2 Answers 2

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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.

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$$\sec\theta+\tan\theta=2\cos\theta=\frac2{\sec\theta}$$

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

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

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

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

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), $\theta=(2m+1)\pi-\frac\pi6\implies m=0$

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    @MSC, try to avoid division & multiplication or squaring which often introduce extraneous roots which needs elimination2013-05-12