Question

How do you solve `2cosxsinx-cosx=0`?

Answer 1

`x={(2k+1)pi/2 ,kin ZZ}uu{kpi+(-1)^k*pi/6 ,kinZZ}`

Explanation:

Here,

`2cosxsinx-cosx=0`

`=>cosx(2sinx-1)=0`

`=>cosx=0 or 2sinx-1=0`

`=>cosx=0 or sinx=1/2`

`(i)cosx=0=>color(blue)(x=(2k+1)pi/2 ,kin ZZ`

`(ii)sinx=1/2=sin(pi/6)`

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

So, the general solution of eqn. is :

`x={(2k+1)pi/2 ,kin ZZ}uu{kpi+(-1)^k*pi/6 ,kinZZ}`
…………………………………………………………………………………………

Note: If `x in[0,2pi) ,or x in[0,360^circ) ,then`

`(I)cosx=0=>x=pi/2,(3pi)/2to{or90^circ,270^circ}`

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

`=>x=pi/6,(5pi)/6to{ or30^circ,150^circ}`

Answer 2

`pi/2 + pi n " and "`

`pi/6 + 2pi n " and "(5 pi)/6 + 2 pi n`

Explanation:

Given: Solve `2 cos x sin x - cos x = 0`

Factor first:

`2 cos x sin x - cos x = cos x (2 sin x - 1) = 0`

`cos x = 1 " and " 2 sin x - 1 = 0`

`pi/2; (3pi)/2 " and " sin x = 1/2`

`pi/2; (3pi)/2 " and " pi/6, (5 pi)/6`

Since an interval isn't given the answer needs to be all values.

`pi/2 " and " (3pi)/2` are `pi` away from each other, so we only need to give one answer:

`pi/2 + pi n`, where `n` is an integer

The other angle values are:

`pi/6 + 2pi n " and "(5 pi)/6 + 2 pi n`