document.write( "Question 596436: find all solutions of the equation in the interval [0,2pi] algebraically\r
\n" ); document.write( "\n" ); document.write( "2sin^2x +3cosx=3 \n" ); document.write( "

Algebra.Com's Answer #377754 by jsmallt9(3758) $\"\"$ $\"About$

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Solving these problems can be broken down into three stages:

Using Trig properties and/or algebra, transform the given equation into one or more equations of the form:
\n" ); document.write( "TrigFunction(variable-expression) = some-number
Find the general solution for the equation(s) from part 1
If the problem asks for solutions in a specific interval, like yours, use the general solution (step 2) to find the specific solution(s) in the given interval.

\n" ); document.write( " $\"2sin%5E2%28x%29+%2B3cos%28x%29=3\"$
\n" ); document.write( "Try as we might, there is no way to achieve the desired form using just algebra. So we'll need to use one or more Trig properties to replace one or more parts of the equation. The $\"sin%5E2%28x%29\"$ suggests using the Pythagorean property (which also has $\"sin%5E2%28x%29\"$ ). Replacing $\"sin%5E2%28x%29\"$ with $\"1-cos%5E2%28x%29\"$ we get:
\n" ); document.write( " $\"2%281-cos%5E2%28x%29%29+%2B3cos%28x%29=3\"$
\n" ); document.write( "Notice the use of parentheses. It is an extremely good habit to use parentheses when substituting one expression for another. In this case the parentheses remind us that we should distribute the 2 in front. Simplifying we get:
\n" ); document.write( " $\"2-2cos%5E2%28x%29+%2B3cos%28x%29=3\"$
\n" ); document.write( "This is a quadratic equation. To solve for cox(x) we need one side to be zero. Subtracting 3 from each side we get:
\n" ); document.write( " $\"-1-2cos%5E2%28x%29+%2B3cos%28x%29-3=0\"$
\n" ); document.write( "then we factor. To make the factoring easier, I'm going to rearrange the terms:
\n" ); document.write( " $\"-2cos%5E2%28x%29+%2B3cos%28x%29-1=0\"$
\n" ); document.write( "and then multiply both sides by -1 (the expression will be easier to factor if it doesn't start with a negative):
\n" ); document.write( " $\"2cos%5E2%28x%29-3cos%28x%29%2B1=0\"$
\n" ); document.write( "Factoring we get:
\n" ); document.write( "(2cos(x) - 1)(cos(x)-1) = 0
\n" ); document.write( "(If you have trouble seeing how this factoring was done, it may help to think of cos(x) as some variable. Let's say cos(x) = q. Then $\"2cos%5E2%28x%29-3cos%28x%29%2B1=0\"$ becomes $\"2q%5E2+-3q+%2B+1+=+0\"$ . You should be able to see that this factors into (2q-1)(q-1) = 0. Then just put cos(x) back in for q.)

\n" ); document.write( "Now we can use the Zero Product Property which tells us that this (or any product) can be zero only if one (or more) of the factors is zero. So:
\n" ); document.write( "2cos(x) - 1 = 0 or cos(x)-1 = 0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "cos(x) = 1/2 or cos(x) = 1
\n" ); document.write( "After all this we have finally transformed the original equation into the desired form. (Fortunately this is usually the longest, hardest part of these problems.)

\n" ); document.write( "Now we find the general solution. Cos's of 1/2 and 1 should be recognizable. Only special angles have a cos of 1/2 or 1. For a cos of 1/2, the reference angle must be $\"pi%2F3\"$ and the angle must terminate in the 1st or 4th quadrant. This gives us:
\n" ); document.write( " $\"x+=+pi%2F3+%2B+2%2Api%2An\"$
\n" ); document.write( "or
\n" ); document.write( " $\"x+=+-pi%2F3+%2B+2%2Api%2An\"$
\n" ); document.write( "And for cos(x) = 1 there is only one angle, 0, so:
\n" ); document.write( " $\"x+=+0+%2B+2%2Api%2An\"$
\n" ); document.write( "(Remember, all angles co-terminal with $\"pi%2F3\"$ , $\"-pi%2F3\"$ or 0 will also work. That is what the \"+ $\"2%2Api%2An\"$ \" is about. This is how we can express all the co-terminal angles mathematically. The \"n\" can be any integer and by using different values for n you get different but co-terminal angles.)

\n" ); document.write( "So the general solution is:
\n" ); document.write( " $\"x+=+pi%2F3+%2B+2%2Api%2An\"$
\n" ); document.write( "or
\n" ); document.write( " $\"x+=+-pi%2F3+%2B+2%2Api%2An\"$
\n" ); document.write( "or
\n" ); document.write( " $\"x+=+0+%2B+2%2Api%2An\"$

\n" ); document.write( "Last of all we find the specific solution(s) that are in the interval [0, $\"2pi\"$ ). For this we take each equation and try different integer values for n until you have found all the solutions in the desired interval. It just so happens that you get only one solution in the interval from each of the three equations. (Note: Sometimes you might get more than one solution in the interval from an equation so don't just stop as soon as you find one. Also, there might be times when an equation from the general solution will not give you any solutions in the interval.)

\n" ); document.write( "When n = 0 in the first equation you get $\"x+=+pi%2F3\"$
\n" ); document.write( "When n = 1 in the second equation you get $\"x+=+5pi%2F3\"$
\n" ); document.write( "When n = 0 in the third equation you get $\"x+=+0\"$
\n" ); document.write( "(Note: With the \")\" next to $\"2pi\"$ in the interval, it means \"up to but not including $\"2pi\"$ . If it had been \"]\", then $\"2pi\"$ would be an acceptable solution and the third equation would give us 2 solutions: for n = 0 and for n = 1) \n" ); document.write( "

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