find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi Thanks a bunch

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Originally Posted by Luckegrl_16 find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi Thanks a bunch $\displaystyle 2sin(2x) = -1$ $\displaystyle sin(2x) = -\frac{1}{2}$ $\displaystyle 2x = \frac{7\pi}{6}, \frac{11\pi}{6}$ So $\displaystyle x = \frac{7\pi}{12}, \frac{11\pi}{12}$ -Dan

oh and srry about the whole multiple posts thing, I've never really used these b4, but now I know