y= 2sin^2 x
y=1- sinx
find values of x inthe interval 0<x<360
if
2sin^x = 1-sinx
this can be arranged into the quadratic.
2sin^2 x + sin x -1=0
(2sinx -1)(sinx + 1)=0
or sinx = 1/2 and sin x=-1
To find the values of x in the interval 0 < x < 360 for the equation 2sin^2(x) = 1 - sin(x), we need to solve the quadratic equation 2sin^2(x) + sin(x) - 1 = 0.
Let's factor the quadratic equation:
(2sin(x) - 1)(sin(x) + 1) = 0
Now, set each factor equal to zero and solve for sin(x):
1) 2sin(x) - 1 = 0
2sin(x) = 1
sin(x) = 1/2
2) sin(x) + 1 = 0
sin(x) = -1
Now, we need to find the values of x for each of these sin(x) values in the interval 0 < x < 360.
1) sin(x) = 1/2
To find the values of x, we can use the inverse sine function (sin^-1) or a calculator. The inverse sine function returns the angle whose sine is equal to a given value.
Using a calculator, sin^-1(1/2) ≈ 30 degrees or π/6 radians.
So, one solution is x = 30 degrees or π/6 radians.
Additionally, sin(x) = 1/2 repeats every 180 degrees or π radians. So, we add multiples of 180 degrees or π radians to get more solutions in the given interval:
x = 30 degrees or π/6 radians + k * 180 degrees or k * π radians, where k is an integer other than 0.
2) sin(x) = -1
Again, using the inverse sine function, sin^-1(-1) = -90 degrees or -π/2 radians.
So, one solution is x = -90 degrees or -π/2 radians.
Similarly, sin(x) = -1 repeats every 180 degrees or π radians, so we can add multiples of 180 degrees or π radians to get more solutions:
x = -90 degrees or -π/2 radians + k * 180 degrees or k * π radians, where k is an integer.
Therefore, the values of x in the interval 0 < x < 360 for the equation 2sin^2(x) = 1 - sin(x) are:
x = 30 degrees or π/6 radians + k * 180 degrees or k * π radians, where k is an integer, and
x = -90 degrees or -π/2 radians + k * 180 degrees or k * π radians, where k is an integer.