Just to make it easier to see (and write), let's call sin(x) 'S' for a while.
2 S2 = 1 - S
Add S to each side ==> ( 2 S2 + S = 1 )
Subtract 1 from each side:
2 S2 + S - 1 = 0
This is just a plain old quadratic equation, which is easy to factor and solve.
(2S - 1) (S + 1) = 0
2S - 1 = 0
2S = 1
S = 0.5
S + 1 = 0
S = -1
Those are the two solutions for 'S'.
Now it's time to recall that 'S' is really sin(x).
sin(x) = 0.5
x = 30° and 150°
sin(x) = -1
x = 270°
Sin squared is equal to 1 - cos squared.
sin squared
Sin squared, cos squared...you removed the x in the equation.
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
Do sin(x), square it, and then multiply it by two.
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.
sin(x)*sin2(x) = 1 so sin3(x) = 1 so that sin(x) = cuberoot(1) = 1 then x = pi/2 + n*pi where n is an integer.
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
2 x cosine squared x -1 which also equals cos (2x)