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cos^2 x + sinx+1 = 0
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\[\text{ use } \cos^2(x)=1-\sin^2(x)\]
Put it in terms of one trig function By using the identity I mentioned above you can do this
o okay so all cosine
\[\text{ since } \cos^2(x)=1-\sin^2(x) \text{ then you can replace } \cos^2(x) \text{ with }\] \[1-\sin^2(x)\]
So it will be in terms of sine not cosine
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oh ok can u help me with something else while ur here
\[1-\sin^2(x)+\sin(x)+1=0\] Do you see that I just replaced \[\cos^2(x) \text{ with } 1-\sin^2(x)\]
yaaa i remmeber tha rule from class
ok so if i have cos x = -0.438
|dw:1339368496399:dw|
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So this is a different question?
is that the reference angle?
ya
So you want to find x right?
wait so do i type it in calc and that ALWAYS gives me the angle in quadrant 1?
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