Question

How do you find the exact value of `sqrt3-sinx=sinx+sqrt12` in the interval `0<=x<360`?

Answer 1

`" "x=-pi/3" " or" "x=(4pi)/3`

Explanation:

`" "`
Solving the equation by transferring the knowns to one side and
`" "`
unknown to the second side.
`" "`
In this equation we will use the following identities:
`" "`
`color(blue)(sin(-alpha)= - sinalpha)`
`" "`
`color(red)(sin(pi+alpha)=-sinalpha)`
`" "`
`-sinx - sinx =-sqrt3+sqrt12`
`" "`
`rArr-2sinx=-sqrt3+sqrt(2^2xx3)`
`" "`
`rArr-2sinx = -sqrt3+2sqrt3`
`" "`
`rArr-2sinx = +sqrt3`
`" "`
`rArrsinx = -sqrt3/2`
`" "`
`rArrsinx = - sin(pi/3)`
`" "`
`sin<0" "` therefore the angle `in`Quadrant 3 or 4

`rArrsinx = color(blue)(sin(-pi/3))" " Or " "sinx = color(red)sin(pi+pi/3)`
`" "`
Hence ,`" "x=-pi/3" " or" "x=(4pi)/3`

Answer 2

`240^@ and 300^@`

Explanation:

sqrt3 - sin x = sin x + sqrt12
After transposing:
`- 2sin x = sqrt12 - sqrt3 = 2sqrt3 - sqrt3 = sqrt3`
`sin x = - sqrt3/2`
Trig table of special arcs and unit circle give 2 solutions:
`sin x = - sqrt3/2` -->
arc `x = - pi/3 = - 60^@` and arc `x = - (2pi)/3 = - 120^@`
The arcs `240^@` and `300^@` are co-terminal to the arcs `- 120^@` and `- 60^@`

Answers for (0, 360)
`240^@ and 300^@`