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An equation is given. Round terms to three decimal places where appropriate. sin(2theta) = 6cos(2theta) Find the solutions in the interval [0, 2pi).
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theta = (1/2)*arctan(6)
2sinxcosx=6(cos^2x-sin^2x) 3cotc-3tanx=1 3-3tan^2x=tanx 3u2+u-3=0 u=tanx u=-1+-sq(37)]/6=tanx tanx=-1.18 x=pi-.145 , 2pi-.145 x=703 , pi+.703 pi=3.14
Just use the fact that sinx/cosx=tan x I think that should give if you devide both sides by cos(2theta) than equation becomes tan(2theta)=6
yeap tanx
2x= arctan6 x=arctan6/2 am i right?
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