# How do I write down the number of solution of the equation of Sin2x-cosx=0 for 0≤x≤360 ?

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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

sol:-

`-cos(x) + 2 cos (x) sin(x)=0`

`cos(x) [2sin(x)-1]=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 .

2sinxcosx-cosx=0

(2sinx-1)cosx=0

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