Question 234142
SOLVE EACH EQUATION, WHERE 0 DEGREES (< OR = TO) X < 360 DEGREES. ROUND APPROXIMATE SOLUTIONS TO THE NEAREST TENTH OF A DEGREE 3 SIN X - 5 + CSC X = 0
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{{{3*Sin(X) - 5 + Csc(X) = 0}}}

Change {{{Csc(X)}}} to {{{1/Sin(X)}}}

{{{3*Sin(X) - 5 + 1/Sin(X) = 0}}}

Multiply through by {{{Sin(X)}}} to clear of fractions:

{{{3*Sin(X)*Sin(X) - 5*Sin(X) + (1/Sin(X))*Sin(X) = 0*Sin(X)}}}

{{{3*Sin(X)*Sin(X) - 5*Sin(X) + (1/cross(Sin(X)))*cross(Sin(X)) = 0*Sin(X)}}}

{{{3*Sin^2(X) - 5*Sin(X) + 1 = 0}}}

{{{3*Sin^2(X) - 5*Sin(X) + 1 = 0}}}

Use the quadratic formula:

where {{{x=Sin(X)}}}, {{{a = 3}}}, {{{b=-5}}}, and {{{c=1}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{Sin(X) = (-(-5) +- sqrt( (-5)^2-4*(3)*(1) ))/(2*(3)) }}}

{{{Sin(X) = (5 +- sqrt(25-12 ))/6 }}}

{{{Sin(X) = (5 +- sqrt(13))/6 }}}

Using the +

{{{Sin(X) = (5 + sqrt(13))/6 = 1.434258546}}}

That's impossible because all sines are between -1 and +1.

Using the -

{{{Sin(X) = (5 - sqrt(13))/6 = 0.2324081208}}}

That is a positive number and is between -1 and +1,
so that will give us two solutions, since the sine
is positive in quadrant I and quadrant II.

To find the quadrant I solution, we find the
inverse sine of 0.2324081208 which is
13.56526739° which rounds to 13.6° to the nearest
tenth of a degree.

That's the reference angle, and in Quadrant I, the
solution IS the reference angle, 13.6°

To find the quadrant II solution, we subtract the
reference angle, 13.6° from 180° and get 166.4°

So the two solutions are X = 13.6° and X = 166.4°.

Edwin</pre>