Question 261029
Since the equation you posted has a terribly messy solution, I am going to assume that the equation is actually {{{2sin^2(x)-3sin(x) = -1}}}<br>
Would you know how to solve {{{2q^2 - 3q = -1}}}? I ask because solving your equation involves<ul><li>Knowing how to solve my equation; and</li><li>Recognizing that your equation has the same structure as my equation.</li></ul>
First let's look at my equation. It is a quadratic equation. So we solve it by getting one side equal to zero:
{{{2q^2 - 3q + 1 = 0}}}
and then factoring the left side (or using the Quadratic Formula). This factors fairly easily:
{{{(2q - 1)(q - 1) = 0}}}
By the Zero Product Property, this product is zero only if one of the factors is zero:
{{{2q - 1 = 0}}} or {{{q - 1 = 0}}}
Solving these we get:
{{{q = 1/2}}} or {{{q = 1}}}<br>
Next you need to see that your equation and mine have the same structure. They both say: "2 times something squared minus 3 times that something equals -1". If you still have trouble seeing this, then use a temporary variable:
Let q = sin(x)
Then substitute this variable into your equation for sin(x):
{{{2(q)^2 - 3*(q) = -1}}}
which is my equation! So you will end up with my solution:
{{{q = 1/2}}} or {{{q = 1}}}
Of course your are trying to solve for x, not the temporary variable q. So substitute sin(x) back in for q:
{{{sin(x) = 1/2}}} or {{{sin(x) = 1}}}<br>
We still haven't solved for x but we're getting closer. Everything we have done so far is Algebra. At this point we finally need some Trig. We have to know when the sin function is 1/2 and when it is 1. If we know the special angles we know that for sin(x) = 1/2:
{{{x = pi/6 + 2pi*n}}} or {{{x = 5pi/6 + 2pi*n}}}}}}
and for sin(x) = 1:
{{{x = pi/2 + 2pi*n}}}<br>
(Eventually you will learn how to solve problems like this without use of a temporary variable like q. But until then, feel free to use them knowing that at some point you will need to substitute back for it.)