Question

Find all solutions in the interval [0,2π) of the equation.

2sin(x)cos(x)−sin(x) + 2cos(x)−1 = 0.

Answer 1

Explanation:

2sin(x)cos(x) - sin(x) + 2cos(x)-1= 0

<=> sinx( 2cosx -1) + ( 2cosx -1) =0

<=> (sinx +1) + ( 2cosx-1)=0

<=> sinx=-1 or cosx= 1/2

+) sinx=1 => x= pi/2 + k×2pi

+)cosx=1/2 => x= pi/3+ k×2pi or x=-pi/3 + k×2pi

x€ [0,2pi] => x= pi/2, pi/3.