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lgg23
 5 years ago
Solve the following trig equation for all x in the interval [0,2pi):
2 [cos x]^(2) + cos x sin x = 1
lgg23
 5 years ago
Solve the following trig equation for all x in the interval [0,2pi): 2 [cos x]^(2) + cos x sin x = 1

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nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0use doublearc formulas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure I understand the question. so I'm not sure if this is the answer.\[(2\cos ^{2}x1)+cosxsinx=0;2((\cos2x)+cosxsinx=0));2\cos2x+2sinxcosx=0;2\cos2x+\sin2x=0\] if you put in 0 for x, you get 1. If you put \[\pi\] for x, you get 1also. I hope this is the answer you need.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0apply the equation (cos x)^2 +(sin x)^2=1,then get (sin x)^2=(sin x)(cos x)
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