Question

How to solve 3sin(x)=cos^2(x)?

`3sin(x)=2cos^2(x)`?

Answer 1

Given: `3sin(x)=2cos^2(x)`

Divide both sides of the equation by 2:

`3/2sin(x)=cos^2(x)`

Substitute: `cos^2(x) = 1-sin^2(x)`:

`3/2sin(x)=1-sin^2(x)`

Add `sin^2(x)-1` to both sides:

`sin^2(x)+3/2sin(x)-1=0`

Please observe that the above is a quadratic equation where sin(x) is the variable, therefore, we can use the quadratic formula:

`sin(x) = (-3/2+-sqrt((3/2)^2-4(1)(-1)))/2`

`sin(x) = 1/2`

`sin(x) = -2 larr` discard because it is outside of the domain of the sine function.

`sin(x) = 1/2`

`x = sin^-1(1/2)`

The values for this condition are well known:

`x = pi/6` and `x = (5pi)/6`

This repeats at integer multiples of `2pi`:

`x = pi/6+2npi` and `x = (5pi)/6+ 2npi` `n in ZZ`

Answer 2

`x=kpi+(-1)^k*pi/6,kinZZ`

Explanation:

Here,

`3sinx=2cos^2x`

`3sinx=2(1-sin^2x)`

`3sinx=2-2sin^2x`

`2sin^2x+3sinx-2=0`

`2sin^2x+4sinx-sinx-2=0`

`2sinx(sinx+2)-1(sinx+2)=0`

`(2sinx-1)(sinx+2)=0`

`2sinx-1=0 or sinx+2=0`

`2sinx=1 or sinx=-2 !in [-1,1]`

So,

`sinx=1/2`

`sinx=sin(pi/6)`

`x=kpi+(-1)^k*pi/6,kinZZ`

Note:

If `x in[0,2pi),then`

`sinx=1/2 > 0=>I^(st)Quadrant or II^(nd)Quadrant`

`=>x=pi/6 or x=pi-pi/6=(5pi)/6`