Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
What exactly did you "Wolfram"? And, of course, $3\pi/2$ is not a solution of your equation. –  David Mitra Apr 30 '12 at 19:20
    
I think Wolfram simplifies by multiplying by $\cos(\theta)$... –  draks ... Apr 30 '12 at 19:22
    
wolframalpha.com/input/… –  MSC Apr 30 '12 at 19:23
1  
-1 for Wolfram... –  David Mitra Apr 30 '12 at 19:31
4  
@ManiSarkarCallisto: I would not worry unduly about what Wolfram Alpha thinks. –  André Nicolas Apr 30 '12 at 19:55
show 5 more comments

2 Answers

up vote 3 down vote accepted

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.

share|improve this answer
add comment

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

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

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

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

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

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

share|improve this answer
    
@MSC, try to avoid division & multiplication or squaring which often introduce extraneous roots which needs elimination –  lab bhattacharjee May 12 '13 at 7:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.