Question

How do you solve `1+cosx-2sin^2x=0` and find all solutions in the interval `0<=x<360`?

Answer 1

See below

Explanation:

Given `1+cosx-2sin^2x=0`

we can do some changes based on trigonometric identities like

`1+cosx-2(1-cos^2x)=0`

`1+cosx-2+2cos^2x=0` re-ordered we have

`2cos^2x+cosx-1=0` lets be `z=cosx`, then

`2z^2+z-1=0` by quadratic formula

`z=(-1+-sqrt(1+8))/4=(-1+-3)/4` we obtain

`z_1=-1` and `z_2=1/2`

If `z_1=cosx_1=-1` then `x_1=pi=180º`
if `z_2=cosx_2=1/2` then `x_2=pi/3=60º` or `x_2=300º=5pi/3`

Answer 2

`x=60^circ, 180^circ,300^circ`

Explanation:

Here,

`1+cosx-2sin^2x=0`

`=>1+cosx-2(1-cos^2x)=0`

`=>1+cosx-2+2cos^2x=0`

`=>2cos^2x+cosx-1=0`

`=>2cos^2x+2cosx-cosx-1=0`

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

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

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

`=>cosx=-1 or cosx=1/2 ,where, 0^circ <= x <360^circ`

`(i)cosx=-1 < 0=>color(red)(x=180^circ`

`(ii)cosx=1/2 >0=>I^(st)Quadrant or IV^(th)Quadrant`

`:.I^(st)Quadrant to cosx=1/2=>color(red)(x=60^circ`

`IV^(th)Quadrant tocosx=1/2=>color(red)(x=360^circ-60^circ=300^circ`

Hence,

`x=60^circ, 180^circ,300^circ`

Answer 3

`x in{60,180,300}`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)sin^2x+cos^2x=1`

`rArrsin^2x=1-cos^2x`

`1+cosx-2(1-cos^2x)=0`

`2cos^2x+cosx-1=0larrcolor(blue)"in standard form"`

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

`"equate each factor to zero and solve for x"`

`cosx+1=0rArrcosx=-1rArrx=180^@`

`2cosx-1=0rArrcosx=1/2`

`"since "cosx>0" x in first/fourth quadrant"`

`x=cos^-1(1/2)=60^@larrcolor(red)"first quadrant"`

`"or "x=(360-60)^@=300^@larrcolor(red)"fourth quadrant"`

`x in{60,180,300}to0<=x<360`