Trig Equation: Please Help (1 Viewer)

[OH1995]

Solve for 0≤x≤360

sinx=cosx

 

[nightweaver066]

Dividing both sides by cosx

 

[OH1995]

One More: 2sin^2x + cosx = 2

 

[kazemagic]

2sin^2x+cosx=2
2(1-cos^2x)+cosx=2
2-2cos^2x+cosx=2
0=2cos^2x-cosx
0=cosx(2cosx-1)
Therefore
0=cosx or 1/2=cosx

Then you should be able to do the rest

 

[Timske]

2sin^2x + cosx = 2
2(1 - cos^2x) + cosx = 2
2 - 2cos^2x + cosx - 2 = 0
2cos^2x - cosx = 0
cosx(2cosx - 1) = 0
cosx = 0 and cosx = 1/2

 

[Timske]

2sin^2x+cosx=2
2(1-cos^2x)+cosx=2
2-2cos^2x+cosx=2
0=2cos^2x-cosx
0=cosx(2cosx-1)
Therefore
0=cosx or 1/2=cosx

i was too sloz