Question

solve for x in the interval [0,2pi]

2sinx^2+cosx=1

Answer 1

use the identity\[\sin^2x=1-\cos^2x\] then the cosines can be set to zero and factored...

Answer 2

\[2\sin^2x+\cos x=1\]identity:\[\sin^2x=1-\cos^2x\] sub in\[2(1-\cos^2x)+\cos x=1\to 2\cos^2x-\cos x-1=0\]\[(2\cos x+1)(\cos x-1)=0\]\[2\cos x+1=0\to\cos x=-\frac{1}{2}\to x=\frac{2\pi}{3},\frac{4\pi}{3}\]\[\cos x-1=0\to\cos x=1\to x=0,2\pi\]so\[x=\left\{ 0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi \right\}\]

Answer 3

(sin x)^2=1-(cos x)^2
put this in the equation
2-2(cos x)^2+cos x= 1
let cos x =y
2-2y^2+y=1
2y^2-y-1=0
find factor of -2 such that their difference is -1
-2 and 1
2y^2-2y+y-1
2y(y-1)+1(y-1)=0
(2y+1)(y-1)=0
y=-1/2 or y=1
cos x= -1/2 or 1
x= 0, 2pi
cos x=-1/2
x = 2pi/3 or 4pi/3

Answer 4

i got soo close! thanks guys