How do you solve 2sin^2x-sinx=1?

2 Answers

Add a -sinx to both sides to get

2*sin^2x-sinx-sinx=1-sinx

2*sin^2x-2sinx=1-sinx

2sinx(sinx-1)-(1-sinx)=0

2*sinx(sinx-1)+(sinx-1)=0

(sinx-1)*(2*sinx+1)=0

From the last equation we have

sinx=1=>x=1/2 (4*pi*n+pi) where n integer

and

sinx=-1/2 with solutions

x_1 = 1/6 (12*pi*k-pi)

and

x_2 = 1/6 (12*pi*k+7*pi)

where k integer

Oct 31, 2016

x=((7pi)/6+2npi), x=((11pi)/6+2npi) and x=(pi/2+2npi) where n in NN.

Explanation:

2sin^2x-sinx=1

Subtract 1 from each side.#

2sin^2x-sinx-1=0

Factor

(2sinx+1)(sinx-1)=0

Set each factor equal to zero and solve.

2sinx+1=0 and sinx-1=0

sinx=-1/2 and sinx=1

For sinx=-1/2 the unit circle shows that x=(7pi)/6 and x=(11pi)/6 within 0<=x<2pi

=>x=(7pi)/6+2npi and x=(11pi)/6+2npi where n is an integer.

For sinx=1, the unit circle shows that x=pi/2 for 0<=x<2pi

=> x=pi/2+2npi where n is an integer.