2 sin2 x + sin x = 1. Letting s = sin x, we have: 2s2 + s - 1 = (2s - 1)(s + 1) = 0; whence, sin x = ½ or -1, and x = 30° or 150° or 270°. Or, if you prefer, x = π/6 or 5π/6 or 3π/2.
No, (sinx)^2 + (cosx)^2=1 is though
That factors to (a + 1)(a + b) a = -1, -b b = -a
You have to put your heart into it!
The answer is 1. sin^2 x cos^2/sin^2 x 1/cos^2 cos^2 will be cancelled =1 sin^2 also will be cancelled=1 1/1 = 1
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
22
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
2 x cosine squared x -1 which also equals cos (2x)
2 sin2 x + sin x = 1. Letting s = sin x, we have: 2s2 + s - 1 = (2s - 1)(s + 1) = 0; whence, sin x = ½ or -1, and x = 30° or 150° or 270°. Or, if you prefer, x = π/6 or 5π/6 or 3π/2.
Sin squared is equal to 1 - cos squared.
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.
X=1
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.
sin squared
No, (sinx)^2 + (cosx)^2=1 is though