Question

Trig help.

Answer 1

Solve on the interval of 0<x<360.

Answer 2

\[2 \cos ^{2}x=\sin x +1\]

Answer 3

@D3xt3R

Answer 4

First\[sin^2(x)+cos^2(x)=1\Leftrightarrow cos^2(x)=1-sin^2(x)\]Replacing\[2*(1-sin^2(x))=sin(x)+1\]\[2-2sin^2(x)=sin(x)+1\]moving everything to one side\[2sin^2(x)+sin(x)+1-2=0\]\[2sin^2(x)+sin(x)-1=0\]Now i prefer to call sin(x) by y, then i've got\[2y^2+y-1=0\]And we have to solve this function to y, and We've got\[y_1=-1~~and~~y_2=\frac{1}{2}\]Now we have to back, because the problem ask for the value of x

if \[y=-1\]\[sin(x)=-1\]\[x=\frac{3\pi}{2}=270^o\]

if \[y=\frac{1}{2}\]\[sin(x)=\frac{1}{2}\]\[x=\frac{\pi}{6}=60^o~~or~~x=\frac{5\pi}{6}=150^o\]The answer is\[\therefore S=\left\{\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\right\}\]

Answer 5

OMG, that was so easy!!!! (once you showed me the light!)

Answer 6

;D
It's normal.

Answer 7

nothing about this is normal. lol. I have one more to post. I'll tag you if it's ok.

Answer 8

No worry, yes you can tag me there ;)