Question

solve the equation 2sin^(2)x=2+cosx

Answer 1

i was thinkng plugin 1-cos^2 but theres no theta symbol

Answer 2

2sin^2(x) = 2 + cos(x)
2sin^2(x) = 1-cos(2x)
cos(x)+cos(2 x)+1 = 0
cos(x)+cos(2x) = -1
x = pi/2, 2pi/3, 4pi/3, 3pi/2 for 0 <= x < 2pi

Answer 3

2(1-cos^2x) = 2 + cosx

cosx + cos^2x = 0

x = pi,pi/2

Answer 4

2*sin^2x=2+cosx
or, 2* (1-cos^x)=2+cosx (since sin^2x+cos^2x=1)
or 2-2*cos^2x=2+cosx
or 2*cos^2x+cosx=0
or cosx (2cosx+1)=0
thus cosx=0 or cosx=-1/2

so, x=pi/2 or 5*pi/3

Answer 5

wait guys the x isnt part of the the exponent

Answer 6

its sin^2 seperate then x