Question 615276
{{{cos^2(x)=1-sin(x)}}}
First use {{{cos^2(x) = 1 - sin^2(x)}}} to change the equation into an "all-sin" equation:
{{{1 - sin^2(x)=1-sin(x)}}}<br>
Next we solve for sin(x). This equation is quadratic in terms of sin(x) so we want one side to be zero. Subtracting 1 and adding {{{sin^2(x)}}} to each side we get:
{{{0 = sin^2(x) - sin(x)}}}
Factor. The GCF is sin(x):
0 = sin(x)(sin(x) - 1)
Use the Zero Product Property:
sin(x) = 0 or sin(x) - 1 = 0
Solving the second equation we get:
sin(x) = 0 or sin(x) = 1<br>
Now that we have solved for sin(x), we can solve for x. Sin values of 0 and 1 are special angle values. So we do not need a calculator. From the special angles we know that...
For sin(x) = 0:
{{{x = 0 + 2*pi*n}}}
or
{{{x = pi + 2*pi*n}}}
For sin(x) = 1:
{{{x = pi/2 + 2*pi*n}}}<br>
From these three equations we can find all the x's that are in the specified interval.
From {{{x = 0 + 2*pi*n}}}...
We get no solutions. (Did you post the interval correctly? I ask because often there is a bracket, not a parenthesis, on one or maybe both ends of the interval. A bracket at one of end of an interval means "including the number at that end". A parenthesis means "not including...." Since you posted parentheses at both ends, 0 and {{{2pi}}} are excluded. If either one of these is supposed to be included, then this equation says they would be a solution in the interval.)<br>
From {{{x = pi + 2*pi*n}}}...
we get {{{x = pi}}} as the only solution within the specified interval.<br>
From {{{x = pi/2 + 2*pi*n}}}...
we get {{{x = pi/2}}} as the only solution within the specified interval.<br>