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Solve the equation for 0 to 360 degrees. (square root of 2)*tan(x) = 2*sin(x)
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\[\sqrt {2} \tan x = 2 \sin x\]\[\sqrt {2} \frac {\sin x}{\cos x} = 2 \sin x\]\[\sqrt {2} \sin x = 2 \sin x \cos x, a \sin u \cos u = \frac {a}{2} \sin (2u)\]\[\sqrt {2} \sin x = \sin 2x\]
And I don't know what to do after that! :(
If you use a graphing calculator, you could see that sqrt (2)sin x and sin (2x) intersect at every pi interval, so for any integer n, the general solutions to equations are\[x = \pi n\]On the interval 0 to 360 degrees, the solutions are \[x = 0, \pi, 2 \pi\]Which in degrees would be 0, 180, 360.
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