find all possible solutions where 0=x=2pi. 1)cot(x)=v3. 2) sin^2x-1=0. 3) cosx-xos^3x=0. 4)cos(2x)=-v3/2.

Question

find all possible solutions where 0≤x≤2pi. 1)cot(x)=√3. 2) sin^2x-1=0. 3) cosx-xos^3x=0. 4)cos(2x)=-√3/2. 5)sin(3x)=-1. 7)cos^2x=sin^2(x/2). i would be really helpful if i can see steps.I got those wrong on my test and i want to see why.Thanks

Answer 1

1)cot(x)=√3.  Equivalently, tan x = 1/√3.

Whenever you see square root of three, you're dealing with a (reference) triangle of 30-60-90 angular composition.

In this case, x = pi/6, a.k.a. 30^o.  Placing the reference triangle in Quadrant III, where tangent is also positive, we get x = 210^o or 7pi/6 radians.

 

2) sin^2x-1=0. Equivalently, sin²x = 1^...NOT sin(2x).

Therefore sin x = ±1 after taking square root of both sides.

That gives x = pi/2 or 3pi/2 as solutions.

 

3) this equation is garbled--you should retype it carefully. (If you mean cos x - cos³x=0, you can solve that by factoring.

 

4) cos(2x)=-v3/2. So argument 2x must be 5pi/6...i.e. a 30-60-90 reference triangle in Quadrant II. Or 7pi/6, Quadrant III, where cosines are also negative.

 

Get the general idea? (The key step is to place an appropriate reference triangle, of the 30-60-90 kind, in the proper position.)