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

2 Answers
Nov 25, 2016

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

Explanation:

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

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

Nov 25, 2016

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^@