How do you solve for x for this equation: 2sin(2x-60°)=1? where -2π ≤ x ≤ 2π [π=pi] I am unsure of where to start for this one
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Originally Posted by Jd09 How do you solve for x for this question: 2sin(2x-60°)=1? where -2π ≤ x ≤ 2π I am unsure of where to start for this one $\displaystyle 2 \sin (2x - 60) = 1 \implies \sin (2x - 60) = \frac 12 $ $\displaystyle \implies 2x - 60 = \sin^{-1} \left( \frac 12 \right) \implies x = \frac {\sin^{-1}(1/2) + 60}2$ so now what? what are all the solutions? which have x in the given range?
Is this correct? (sin-1)sin(2x-60)=1/2(sin-1) (2x-60)=30 x=45 ??
Originally Posted by Jd09 Is this correct? (sin-1)sin(2x-60)=1/2(sin-1) (2x-60)=30 x=45 ?? that is correct...is that the only solution? you want $\displaystyle -360^o \le x \le 360^o$ remember?
135... and would -135 work as well?
Thank you for your help!
Last edited by mr fantastic; Apr 7th 2009 at 10:07 PM. Reason: Moved the rest of the post to a new thread
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