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How do I solve: tan^2(x)-3tan(x)+2=0
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let tanx=u then it's u^2-3u+2=0 or u = 1, 2 or tanx=1 => x = pi/4 or tanx = 2 => x= arctan(2)
let tan x =A A^2-3A+2=0 now solve it??
tan^2(x)-3tan(x)+2=0 (tan(x)-2)(tan(x)-1)=0 tanx=2 , tanx=1 x=45, 63.43 , 243.43 , 225
it would probably help if you let tan(x) = u u^2 -3u + 2 = 0 solve the quadratic function..then substitute u back to tan(x)..arctangent the values to get x :DDD
thanks guys, I get tanx=1 equals pi/4 and 5pi/4 because of the unit circle but how would I find the values of tanx=2 on that?
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tanx=2 must use calculator
Alright I really appreciate everyone's help! you're are the best :)
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