Sin2x+cosx=0 solve the equation on the interval 0,2pi?

1 Answer
May 14, 2018

x=pi/2, (7pi)/6, (3pi)/2, (11pi)/6

Explanation:

Recall the identity sin2x=2sinxcosx. Rewrite the equation with this identity:

2sinxcosx+cosx=0

And note that we can factor out cosx:

cosx(2sinx+1)=0

We now solve the equations cosx=0, 2sinx+1=0:

cosx=0 -> x=pi/2, (3pi)/2 solve this on the interval [0, 2pi).

2sinx+1=0

2sinx=-1

sinx=-1/2

x=(7pi)/6, (11pi)/6 solve this on [0, 2pi).