Equation Solving: solve 2 sin x + 1 = 0 for x, 0 <= x < 2pi

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
hakandragon
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Equation Solving: solve 2 sin x + 1 = 0 for x, 0 <= x < 2pi

Postby hakandragon » Mon Aug 17, 2009 3:18 am

If someone could point me in the right direction of where to go with these sorts of problems, I would be so greatful :p

Solve the equation for x, where 0</x<2pi

2 sin x + 1 = 0

:confused:

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stapel_eliz
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Postby stapel_eliz » Mon Aug 17, 2009 1:17 pm

How would you solve 2x + 1 = 0? Start the solution in the same manner.

For what value of x is sin(x) equal to -1/2? Complete the solution using this information. :wink:

hakandragon
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Re: Equation Solving: solve 2 sin x + 1 = 0 for x, 0 <= x < 2pi

Postby hakandragon » Mon Aug 17, 2009 4:35 pm

I'm sorry, I know this is probably a problem I should be able to do in my head or something, but I'm still confused :(

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stapel_eliz
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Postby stapel_eliz » Mon Aug 17, 2009 5:13 pm

Okay, so first you'll need to re-learn how to solve linear equations. Once you've got that down, you can solve for "sin(x)=".

Then you'll need to re-learn the basic reference-angle values for the trig functions. Once you've got that down, you can find at least one solution to the trig equation. To find the other solution, you'll need to review the sine values for one period.


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