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How to solve trigonomtry function involving $\sin x \cos x$ and $\sin 2x$:

$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$

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You last three lines should be removed. "help me," "thank you" etc should not be part of the question. The way you can show appreciation is by voting answers (good: up & poor: down), accepting an answer, replying to clarification comments, and most importantly, showing what you tried & where you got stuck. –  user2468 Aug 17 '12 at 15:37
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@JenniferDylan Ok, but politeness should never be discouraged :-) –  Siminore Aug 17 '12 at 15:53
    
After your several questions in trigonometry, it'd be nice if you showed some self-work, some insights, ideas, effort in solving them. –  DonAntonio Aug 17 '12 at 16:29

1 Answer 1

up vote 4 down vote accepted

Hint:

Using the identity $\sin(2x) = 2 \sin x \cos x$ we have $$ \sin x \cos x + \sin x + 2\cos x + 2 = 0$$ Factor $$ (1 + \cos x) \sin x + 2(1 + \cos x) = 0 \\ (1 + \cos x)(2 + \sin x) = 0 $$ So either $1 + \cos x = 0$ or $2 + \sin x = 0.$ Solve for $x$ in each case.

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the answer is x = 0 rad and 2 pi rad –  dramasea Aug 17 '12 at 15:42
    
@dramasea No, that's not correct. First, $\sin x = -2$ has no real solutions. Why? Well because $\sin x \in [-1, 1].$ Second case: if $\cos x = -1,$ then what is $x$? Draw the unit circle and think about it. Also, please note that if $\theta$ is a solution, then $\theta + 2\pi, \theta + 4\pi, \theta + 6\pi, \dots$ are also solutions. –  user2468 Aug 17 '12 at 15:49
    
thanks!!! can u help me solve $$\tan2x-\sin4x = 0$$ –  dramasea Aug 17 '12 at 15:50
    
@dramasea it's a customary of math.SE to post separate threads for different questions, so please go ahead & post a new question. I (or other users) will try to help you. –  user2468 Aug 17 '12 at 15:51
    
here's the link math.stackexchange.com/questions/183694/… tq –  dramasea Aug 17 '12 at 15:58

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