solve the trigonometric equation sin 2x-sin x +2cos x -1=0 in the interval [0,2TT)

Question

alexwee123 · Answer

is this the equation?
$$\sin ^{2}x-sinx+2cosx-1=0?$$ or
$$\sin2x-sinx+2cosx-1=0?$$

dumbcow · Answer

its most likely the sin(2x) option
its easier and solutions are easily in terms of pi

alexwee123 · Answer

ugh i didn't do trig identities in a while so....
$$\sin2x=2sinxcosx$$
so...
$$2sinxcosx+2cosx=sinx+1$$$$2cosx(sinx+1)=sinx+1$$$$2cosx=1$$$$cosx=1/2$$$$\cos ^{-1}(\cos(x))=\cos ^{-1}(1/2)$$$$x=\Pi/3$$
did i make a mistake @dumbcow ? o.0

anonymous · Answer

sinx+1=1?
only when squared

dumbcow · Answer

hmm i didn't see that solution but i see no flaw in your work

anonymous · Answer

wait its sin^2 + cos^2 = 1

alexwee123 · Answer

ya ^^

anonymous · Answer

i wish i knew the answer to this problem its gonna bug me lol

alexwee123 · Answer

@dumbcow what did you get as an answer?
sorry that kind of sounded rude :/

anonymous · Answer

pie/3, 5pie/3 is what i got

dumbcow · Answer

no its fine...i went back and realized i did it wrong
you are correct but there are other solutions...instead of cancelling out the sinx +1
factor it out
$$\rightarrow (\sin x +1)(2\cos x -1) = 0$$
$$x = \pi/3 , 3\pi/2, 5\pi/3$$

alexwee123 · Answer

lol ya i see now :/

alexwee123 · Answer

but 3pi/2 is extraneous

anonymous · Answer

i dont think 3pie/2 is part of the answer