How do I write down the number of solution of the equation of Sin2x-cosx=0 for 0≤x≤360 ?
I suppose that you have to completely solve the equation sin(2x) - cos(x) = 0. Correct me If you mean something else.
Express sin(2x) as 2*sin(x)*cos(x). So the equation may be rewritten as
2*sin(x)*cos(x) = cos(x), or cos(x)*(1-2*sin(x)) = 0.
So cos(x) = 0 or 1-2*sin(x) = 0.
The first variant, cos(x)=0, gives us x1 = 90°, x2 = 270° (remember we are consider only 0°<=x<=360°).
The second variant, sin(x)=1/2, gives us x3 = 30° and x4=150°.
The number of solutions is 4.
To find the number of Solutions of the equation
Sin2x-cosx=0 for 0≤x≤360
`-cos(x) + 2 cos (x) sin(x)=0`
`=> cos(x) =0 or (2sin(x) -1) =0`
` cos(x) =0 , 0<=x<=2pi `
`: x=pi/2 , x=3*pi/2`
` 2sin(x) -1 =0 , 0<=x<=2pi`
` : x= pi/6 , x =(5*pi)/6`
` x=pi/2 , x=3*pi/2 ,x= pi/6 , x =(5*pi)/6`
are the four possible solutions for the given equation .
in the interval [0,360] we get cosx=0 at x= `pi/2,(3pi)/2`
and sinx=1/2 at x=`pi/6,(5pi)/6`
so we have 4 solutions