Hi All,
Having a bit of trouble with this one, I can find 2 solutions but the answer key is calling for 4.
Find all values of x in the interval [0,2pi] that satisfy
sin2x=cosx
heres how I went about it
sin2x = cosx
2sinxcosx = cosx
then cancel the cos and divide
sinx = 1/2
which gives solutions of pi/6 and 2pi/3. Not sure where the other 2 solutions are coming from as the identity rearranges to this and sin only gives positive answers in 2 quadrants however the answer key is giving me an answer for all 4 quadrants.
dont want the answer but a nudge in the right direction would be super
Having a bit of trouble with this one, I can find 2 solutions but the answer key is calling for 4.
Find all values of x in the interval [0,2pi] that satisfy
sin2x=cosx
heres how I went about it
sin2x = cosx
2sinxcosx = cosx
then cancel the cos and divide
sinx = 1/2
which gives solutions of pi/6 and 2pi/3. Not sure where the other 2 solutions are coming from as the identity rearranges to this and sin only gives positive answers in 2 quadrants however the answer key is giving me an answer for all 4 quadrants.
dont want the answer but a nudge in the right direction would be super