Here's the question you clicked on:
jahvoan
Solve each equation for 0 = x = 2π : i) 2cos^(2)x + 3cosx – 2 = 0 ii) sinx = sqrt 3cosx (Rewards for correct answer)
did you mean \[ (1)\quad2cos^2x+3cosx−2=0\\ (2)\quad \sin x=\sqrt{3}\cos x\\ \text{for }0\le x \le2\pi\] ???
a. (2cosx-1)(cosx+2)=0 cosx=1/2 or cosx=-2 x={π/3, 5π/6} This is correct, i promise...
If you set p=cos x in the first one, you get: 2p²+3p-2=0. You can solve this, using the quadratic formula. Once you've got you p, remember cos x = p, so solve for x.
Do you get mines @jahvoan
i got the answer but what is the corresponding angle for -2 on the unit circle?
There isn't. cosx=-2 has no solutions, because the x-coordinate of a point on the unit circle (which is the cosine) only has values from -1 to 1...
You can divide both sides by cos x and remember sinx/cosx = tan x
but on the right it is sqrt cosx
no look at the problem up top
It says: sinx = sqrt 3cosx This can only mean:\[ \sin x = \sqrt{3} \cdot \cos x\]So dividing by cos x gives:\[\frac{ \sin x }{\cos x }=\sqrt{3} \Leftrightarrow \tan x=\sqrt{3}\]And this is a "nice" value of tan x. If you interpret it otherwise, you can't solve it...
so what will be the answer for sqrt 3 on the unit circle?
Remember that special triangle with sides of a, 2a and a√3?
no its pi/3 right or wrong?
why is equation 2 not considered to be an identity?
Identities hold for every x. This one is only true if x = pi/3. Identity: sinx=cos(pi/2-x). No matter what x is, it's always true!