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Help Solving for X (Easy I think)

  • Thread starter Magnawolf
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  • #1
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sinxcosx+[itex]\frac{1}{2}[/itex]sin2x = 0.0392

I can't figure out how to solve for x on this one...

I tried sin2x = cosxsinx but still can't get anywhere.
 

Answers and Replies

  • #2
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In future posts, don't discard the homework template.

For this problem, I don't see an easy analytic solution that would allow writing the left-hand side in terms of either sin(x) or cos(x). Having said that, there are lots of numerical methods that would give an approximate solution.
 
  • #3
LCKurtz
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sinxcosx+[itex]\frac{1}{2}[/itex]sin2x = 0.0392

I can't figure out how to solve for x on this one...

I tried sin2x = cosxsinx but still can't get anywhere.
Hopefully you mean ##\sin(2x) = 2\sin x\cos x##. Were you given the right side as .0392 or is that a decimal approximation to what you were actually given? Anyway, calling that constant ##k##, you can write your equation as$$
\frac{\sin(2x)}{2} + \frac 1 2(\frac{1-\cos(2x)}{2})=k$$ $$
2\sin(2x) + 1 - \cos(2x) = 4k$$ $$
\frac 2 {\sqrt 5}\sin(2x) - \frac 1 {\sqrt 5}\cos(2x) = \frac {4k-1}{\sqrt 5}$$
You can write that as a single trig function of ##2x## plus a phase angle using an addition formula. You will at least symbolically be able to solve for ##x## although you may need a calculator to evaluate it exactly.
 

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