Question 634796
{{{4cos(2x)-3sin(x)csc^3(x)+6 = 0}}}
As the problem suggest, you want everything expressed in terms of sin(x). SO we will start by substituting for the {{{cos(2x)}}} and {{{csc^3(x)}}}. The csc is easy. csc is the reciprocal of sin. So {{{csc^3(x) = 1/sin^3(x)}}}. We have three variations for cos(2x):<ul><li>{{{cos(2x) = cos^2(x)-sin^2(x)}}}</li><li>{{{cos(2x) = 2cos^2(x)-1}}}</li><li>{{{cos(2x) = 1-2sin^2(x)}}}</li></ul>Since we want everything in terms of sin(x), we will use the third variation. Substituting in we get:
{{{4(1-2sin^2(x))-3sin(x)(1/sin^3(x))+6 = 0}}}
Note the use of parentheses! These are important when making substitutions. Simplifying we get:
{{{4-8sin^2(x)-(3/sin^2(x))+6 = 0}}}
{{{-8sin^2(x)-(3/sin^2(x))+10 = 0}}}<br>
Now that everything is in terms of sin(x), the next task is to get rid of the fraction. Multiplying both sides by {{{sin^2(x)}}} we get:
{{{-8sin^4(x)-3+10sin^2(x) = 0}}}
As you may see, we're getting close. We can get the signs right if we multiply both sides by -1:
{{{8sin^4(x)+3-10sin^2(x) = 0}}}
And rearrange the terms:
{{{8sin^4(x)-10sin^2(x)+3 = 0}}}<br>
Now we can substitute "y" for "sin(x)":
{{{8y^4-10y^2+3 = 0}}}
The reason we made this substitution is that it is easier to see how to solve
{{{8y^4-10y^2+3 = 0}}}
than it is to see how to solve
{{{8sin^4(x)-10sin^2(x)+3 = 0}}}
(Eventually you will see how to solve without making the substitution.) Note: An even better substitution is {{{z = sin^2(x)}}} This would make the substituted equation:
{{{8z^2-10z+3=0}}}
which I think you would agree is even easier to solve than:
{{{8y^4-10y^2+3 = 0}}}
But we will go with the substitution your teacher/book suggested.
<br>
{{{8y^4-10y^2+3 = 0}}}
is an equation in what is called quadratic form. (A "pure" quadratic would have y-squared and y terms.) We can solve it with the same methods used to solve pure quadratics. One of these is factoring:
{{{(4y^2-3)(2y^2-1) = 0}}}
If this factoring is not immediately obvious to you, then look at it for a while and let it sink in. Multiply it back out and see if you get {{{8y^4-10y^2+3 = 0}}}. Look back at our alternate substitution, {{{8z^2-10z+3=0}}}, and think about how that would factor. It might help you see how we got the factors above.<br>
Now we use the Zero Product Property:
{{{4y^2-3 = 0}}} or {{{2y^2-1=0}}}
Solving these...
{{{4y^2 = 3}}} or {{{2y^2= 1}}}
{{{y^2 = 3/4}}} or {{{y^2= 1/2}}}
Finding the square root of each side (and not forgetting about the negative square roots) we get:
{{{y = sqrt(3/4)}}} or {{{y = -sqrt(3/4)}}} or {{{y = sqrt(1/2)}}} or {{{y = -sqrt(1/2)}}}
Rationalizing the denominators...
{{{y = sqrt(3)/sqrt(4)}}} or {{{y = -sqrt(3)/sqrt(4)}}} or {{{y = sqrt((1/2)(2/2))}}} or {{{y = -sqrt((1/2)(2/2))}}}
{{{y = sqrt(3)/2}}} or {{{y = -sqrt(3)/2}}} or {{{y = sqrt(2/4)}}} or {{{y = -sqrt(2/4)}}}
{{{y = sqrt(3)/2}}} or {{{y = -sqrt(3)/2}}} or {{{y = sqrt(2)/sqrt(4)}}} or {{{y = -sqrt(2)/sqrt(4)}}}
{{{y = sqrt(3)/2}}} or {{{y = -sqrt(3)/2}}} or {{{y = sqrt(2)/2}}} or {{{y = -sqrt(2)/2}}}<br>
We've solved for y. But we really want to solve for x. So it is time to substitute sin(x) back in for the y:
{{{sin(x) = sqrt(3)/2}}} or {{{sin(x) = -sqrt(3)/2}}} or {{{sin(x) = sqrt(2)/2}}} or {{{sin(x) = -sqrt(2)/2}}}
All of these are special angle sin's so every x is a special angle (so put your calculator away).<br>
We should recognize that the reference angle for both {{{sin(x) = sqrt(3)/2}}} or {{{sin(x) = -sqrt(3)/2}}} will be {{{pi/3}}}. And since we have both positive and negative {{{sqrt(3)/2}}}, x will terminate in all four quadrants. So for {{{sin(x) = sqrt(3)/2}}} or {{{sin(x) = -sqrt(3)/2}}}:
{{{x = pi/3 + 2pi*n}}} (1st quadrant, {{{sqrt(3)/2}}})
{{{x = pi - pi/3 + 2pi*n}}} (2nd quadrant, {{{sqrt(3)/2}}})
{{{x = pi + pi/3 + 2pi*n}}} (3rd quadrant, {{{-sqrt(3)/2}}})
{{{x = -pi/3 + 2pi*n}}} (4th quadrant, {{{-sqrt(3)/2}}})
with the middle two equations simplifying to:
{{{x = pi/3 + 2pi*n}}} (1st quadrant, {{{sqrt(3)/2}}})
{{{x = 2pi/3 + 2pi*n}}} (2nd quadrant, {{{sqrt(3)/2}}})
{{{x = 4pi/3 + 2pi*n}}} (3rd quadrant, {{{-sqrt(3)/2}}})
{{{x = -pi/3 + 2pi*n}}} (4th quadrant, {{{-sqrt(3)/2}}})<br>
For {{{sin(x) = sqrt(2)/2}}} and {{{sin(x) = -sqrt(2)/2}}} we should recognize that the reference angle will be {{{pi/4}}} and x will again terminate in all four quadrants (since we have both positive and negative {{{ sqrt(2)/2}}}. This gives us:
{{{x = pi/4 + 2pi*n}}} (1st quadrant, {{{sqrt(2)/2}}})
{{{x = pi - pi/4 + 2pi*n}}} (2nd quadrant, {{{sqrt(2)/2}}})
{{{x = pi + pi/4 + 2pi*n}}} (3rd quadrant, {{{-sqrt(2)/2}}})
{{{x = -pi/4 + 2pi*n}}} (4th quadrant, {{{-sqrt(2)/2}}})
with the middle two equations simplifying to:
{{{x = pi/4 + 2pi*n}}} (1st quadrant, {{{sqrt(2)/2}}})
{{{x = 3pi/4 + 2pi*n}}} (2nd quadrant, {{{sqrt(2)/2}}})
{{{x = 5pi/4 + 2pi*n}}} (3rd quadrant, {{{-sqrt(2)/2}}})
{{{x = -pi/4 + 2pi*n}}} (4th quadrant, {{{-sqrt(2)/2}}})<br>
The full general solution to your equation is all 8 of the equations above:
{{{x = pi/3 + 2pi*n}}} (1st quadrant, {{{sqrt(3)/2}}})
{{{x = 2pi/3 + 2pi*n}}} (2nd quadrant, {{{sqrt(3)/2}}})
{{{x = 4pi/3 + 2pi*n}}} (3rd quadrant, {{{-sqrt(3)/2}}})
{{{x = -pi/3 + 2pi*n}}} (4th quadrant, {{{-sqrt(3)/2}}})
{{{x = pi/4 + 2pi*n}}} (1st quadrant, {{{sqrt(2)/2}}})
{{{x = 3pi/4 + 2pi*n}}} (2nd quadrant, {{{sqrt(2)/2}}})
{{{x = 5pi/4 + 2pi*n}}} (3rd quadrant, {{{-sqrt(2)/2}}})
{{{x = -pi/4 + 2pi*n}}} (4th quadrant, {{{-sqrt(2)/2}}})