solve algebriaclly the equation y=2sin2x+1 for 0<_x_<180. when solved find the x coordinates of the roots.
I am assuming you are trying to solve
2sin2x+1 = 0
sin 2x = -1/2
(angle in standard position is 30º)
the sine is negative in quadrants III and IV
so x = 180+30 = 210
or x = 360-x30 = 330
Btw, don't shout!
When you type in all-caps, two things happen:
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my last couple of lines should have been
so 2x = 180+30 = 210
or 2x = 360-x30 = 330
so x = 105 or x = 165
since the period of sin 2x is 180º
two other solutions are 285º and 345º
check: 2sin(2x345) + 1 = 0
To solve the equation algebraically, we need to find the x-coordinates of the roots. Let's begin by simplifying the equation.
The given equation is: y = 2sin^2(x) + 1
First, let's move 1 to the other side of the equation.
2sin^2(x) = y - 1
Now, divide both sides of the equation by 2:
sin^2(x) = (y - 1) / 2
Next, take the square root of both sides of the equation:
sin(x) = ±√((y - 1) / 2)
To solve for x, we need to use the inverse sine function (sin^-1) on both sides
x = sin^-1(±√((y - 1) / 2))
Now, we can find the x-coordinates of the roots by substituting the values of y within the given range (0 ≤ x ≤ 180) into the equation.
For example, if y = 0:
x = sin^-1(±√((0 - 1) / 2))
We will find the roots by substituting other values of y within the given range, such as y = -1, y = 1, etc. By doing this, we can find the corresponding x-coordinates for each root.
Note: Keep in mind that the ± sign indicates that there may be multiple solutions for each value of y.