# If asked to solve the equation:`cosx(2sinx+sqrt(3))(-sqrt(2)cosx+1)=0` for `0<=x<=2Pi` could we simply set each factor equal to zero - giving us the following: `cosx=0 , sinx= -sqrt(3)/2...

If asked to solve the equation:`cosx(2sinx+sqrt(3))(-sqrt(2)cosx+1)=0` for `0<=x<=2Pi` could we simply set each factor equal to zero - giving us the following:

`cosx=0 , sinx= -sqrt(3)/2 ,cosx=1/sqrt(2)`

Then, could we say that the solutions for cos are `cosx = Pi/2 , (3Pi)/(2) , Pi/4 , (7Pi)/(4)` - we get this by looking at our unit circle where `cos = 0 , andcos = (1)/(sqrt(2))` , and then say that our solutions for sin are `sin = (4Pi)/(3) , and (5Pi)/(3)` - we get this by looking at our unit circle where `sin = -(sqrt(3))/(2)` . Would this be wrong to say this? The way the equation is set up makes it very tempting to choose this method in order to solve. If this method of solving is* incorrect*, please explain the *correct* method.

*print*Print*list*Cite

### 1 Answer

The general method is correct.

Solve `cosx(2sinx+sqrt(3))(-sqrt(2)cosx+1)=0`

By the zero product property at least one of the factors must be zero.

cosx=0 ==> `x=pi/2,(3pi)/2`

`2sinx+sqrt(3)=0 ==> sinx=-sqrt(3)/2 ==> x=(4pi)/3,(5pi)/3`

`-sqrt(2) cosx+1=0 ==> cosx=1/sqrt(2)==> x=pi/4,(7pi)/4`

------------------------------------------------------------------

The solutions on `0<=x<=2pi` are `pi/4,pi/2,(4pi)/3,(3pi)/2,(5pi)/3,(7pi)/4`

------------------------------------------------------------------

The graph (horizontal ticks in `pi/12` :