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anonymous
 5 years ago
Can you please explain step by step on how to find the all the solutions to the equation 2cosx(sinx)  cos x=0 in the interval [0,2pi]
Thank you in advance!!!
anonymous
 5 years ago
Can you please explain step by step on how to find the all the solutions to the equation 2cosx(sinx)  cos x=0 in the interval [0,2pi] Thank you in advance!!!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\cos x(2\sin (x)1)=0\] pull out the cos fucntion to get the above expression. So then you have two solution equations. \[\cos(x)=0\] and \[2\sin(x)1 =0\]. So then you are looking for all of the angles where cos = 0 and sin = 1/2. cos = 0 at pi/2 and 3pi /2. I'm pretty sure sin = 1/2 at pi/6, but if not it is pi/3. (30 or 60 degrees). Sin is also positive between 90 and 180 degrees, so there will be a second angle there that equals 1/2. Refer to the unit circle and become ultra familiar with the angles and their values, just because it is better to be able to pull them up off hand. Your answer is thosee four angles.
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