An equation is given. Round terms to three decimal places where appropriate.

sin(2theta) = 6cos(2theta)

Find the solutions in the interval [0, 2pi).

Question

An equation is given. Round terms to three decimal places where appropriate.

sin(2theta) = 6cos(2theta)

Find the solutions in the interval [0, 2pi).

anonymous · Answer

theta = (1/2)*arctan(6)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

yeap
tanx

anonymous · Answer

2x= arctan6

x=arctan6/2

am i right?