Recent content by Liquidice_69

ok thxs so i want to solve this for x over [-2pi, 2pi] cot(x)cos^2(x) = 2 cot(x) cos^2(x) = 2 x= cos^{-1}(\sqrt{2}): but how do i find all of the solution i am looking for from there

but i thought cos(x)^2 doesn't equal cos^2 (x)

im sry if you think I am not trying but I am not taking a trig course I am just in algebra 2 and my teacher wanted to touch on trig. But with the calc you said to type in y= cos^2 x but i don't know to do that with my ti 83+ because when i push the cos button in outputs cos( so i can't put a ^2

ya i know, but i just want to know how to solve cos^2 x = 2 but i don't know how

how do u type cos^2 x = 2 into a graphing calc because when ever use push cos it puts the ( so u can't a ^2 before it.

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ok i get it the answers are for the cos x=1 0 for the cos x=.5 pi/3 and 5 pi/3

i don't get it, do u have to subtract pi from the second equasion

ok, i want to solve it for negative pi to pi

ok here's what i got so far 2sin^2x+3cos x-3 = 0 2(1-cos^2x)+3cos x-3 = 0 -2cos^2x+3cos x-1 = 0 2cos^2x-3cos x+1 = 0 and i know how to factor 2x^2-3x+1=0 (2x-1)(x-1)=0 x=1, x=.5 but how do i solve with the cos

but how do you get the cos^2

im trying to solve this trig equation and the book says to factor it i think but its not clear, could anyone help. thxs 2sin^2 x + 3cos x – 3 = 0