find all solutions of 2sinx=1-2cosx in the interval from 0 to 360
2sinx = 1 - 2cosx
sinx+cosx = 1/2
Now, sin(x+π/4) = sinx cosπ/4 + cosx sinπ/4
= 1/√2 (sinx+cosx)
SO, what you have is just
√2 sin(x+π/4) = 1/2
sin(x+π/4) = 1/√8
x+Ï€/4 = 0.368
x = -0.424
So, on [0,2Ï€) we have
x = 2Ï€-0.424 or (Ï€-0.368)-Ï€/4
x = 5.859 or 1.995
To find the solutions of the equation 2sin(x) = 1 - 2cos(x) in the interval from 0 to 360 degrees, we need to use trigonometric identities and algebraic manipulations.
Let's start by rewriting the equation using trigonometric identities:
2sin(x) = 1 - 2cos(x)
2sin(x) + 2cos(x) = 1
Next, we can rewrite sin(x) and cos(x) in terms of tan(x/2) using half-angle formulas:
2 * (2tan(x/2) / (1 + tan²(x/2))) + 2 * ((1 - tan²(x/2)) / (1 + tan²(x/2))) = 1
Simplifying this equation further, we get:
4tan(x/2) + 2 - 2tan²(x/2) = 1 + tan²(x/2)
3tan²(x/2) - 4tan(x/2) + 1 = 0
Now, let's solve this quadratic equation for tan(x/2). We'll use the quadratic formula:
tan(x/2) = (-b ± √(b² - 4ac)) / (2a),
where a = 3, b = -4, and c = 1. Plugging in these values, we have:
tan(x/2) = (-(-4) ± √((-4)² - 4 * 3 * 1)) / (2 * 3)
tan(x/2) = (4 ± √(16 - 12)) / 6
tan(x/2) = (4 ± √4) / 6
Simplifying further:
tan(x/2) = (4 ± 2) / 6
tan(x/2) = 1 or tan(x/2) = 1/3
Now, to find x, we need to use the inverse tangent (arctan) function:
x = 2 * arctan(tan(x/2))
Therefore, the solutions for x in the interval from 0 to 360 degrees are:
x = 2 * arctan(1) = 2 * 45 = 90 degrees
x = 2 * arctan(1/3) = 2 * 18.43 ≈ 36.87 degrees
So, the solutions are x = 90 degrees and x ≈ 36.87 degrees.