what values for theta(0≤theta≤2pi) satisfy the equation? 2sinthetacostheta+sqrt 3 theta
is there an equation?
yeah
ok so you meant to put =0 at the end
yeah
I'm really confused x.x why r practice packets so hard >:C
and the other cos thingy but yeah anyways you can factor the left hand side \[2 \sin(\theta)\cos(\theta)+\sqrt{3} \cos(\theta)=0 \\ \text{ factor out the } \cos(\theta) \\ \cos(\theta)[2 \sin(\theta)+\sqrt{3}]=0 \]
set both factors =0 and solve for theta
\[\cos(\theta) \text{ or } \sin(\theta) =\frac{-\sqrt{3}}{2}\] use unit circle
??!!
forgot the =0 part :p
\[\cos(\theta)=0 \text{ or } \sin(\theta) =\frac{-\sqrt{3}}{2}\] use unit circle *
??? wits the answer suppose to be :( explanation too pls
sin(theta) is the y-coordinate on the unit circle find for what theta's you have the y value ( the second value) of the pair is -sqrt(3)/2
hint: look in quadrants 3 and 4
okie :( wait the answer is B??
sin(3pi/4) is not -sqrt(3)/2 it is sqrt(2)/2
so theres no answer? ._.
i didn't say that i just said the option you chose is incorrect because sin(3pi/4) is not -sqrt(3)/2
did you look at the unit circle you are looking for when the y-coordinate is -sqrt(3)/2
D
the hint I gave is to look in the 3rd and 4th quadrant
d??
https://www.mathsisfun.com/geometry/images/circle-unit-radians.gif look at this picture do you see the ordered pairs?
look at the second number of each pair you are looking to for which theta you have the y coordinate is -sqrt(3)/2
can you tell me looking at that chart thing where you see y coordinate is -sqrt(3)/2
ok look at that chart do you see under sin -sqrt(3)/2 ?
find it then use your finger and move it horizontally from that number to the theta number
OH I GET WUT MEAN ITS A DUH ITS 4PI/3 IM SO STUPID
yes sin(4pi/3)=-sqrt(3)/2
and sin(5pi/3)=-sqrt(3)/2
and of course cos(theta)=0 is easy to solve and all of the choices included its answers which was pi/2 and 3pi/2
:D thnx so much here a medal
np
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