find the exact solutions (¦È in degrees)
sin^2¦È+2cos¦È=-2
range= 0¡Ü¦È<360¡ã
see the other problem, same procedure.
To find the exact solutions for the equation sin^2 ¦È + 2cos ¦È = -2, we can use trigonometric identities to rewrite the equation in terms of a single trigonometric function.
First, we know that sin^2 ¦È + cos^2 ¦È = 1 (identity 1). We can rewrite the equation by adding sin^2 ¦È and cos^2 ¦È to both sides:
sin^2 ¦È + cos^2 ¦È + 2cos ¦È = -2 + 1
Simplifying, we get:
1 + 2cos ¦È = -1
Next, we rearrange the equation to isolate cos ¦È:
2cos ¦È = -2 - 1
2cos ¦È = -3
Dividing both sides by 2, we get:
cos ¦È = -3/2
Now, we need to find the values of ¦È that satisfy this equation in the given range of 0¡Ü¦È<360¡ã.
To find the exact solutions, we can use the inverse cosine function, or arccos. However, the cosine function only gives outputs in the range of 0¡Ü¦È<180¡ã, so we need to consider the reference angle.
The reference angle is the acute angle between the terminal side of the angle and the x-axis. We can find the reference angle by taking the inverse cosine of the absolute value of the ratio of the adjacent side to the hypotenuse.
In this case, the ratio is |-3/2| = 3/2. Taking the inverse cosine of 3/2, we find the reference angle ¦È_ref:
¦È_ref = arccos(3/2) = 0
Since the cosine function is negative in the second and third quadrants, we can find the other solutions by adding or subtracting the reference angle from a full revolution (360¡ã).
The solutions for ¦È in the given range are:
¦È = 180 - ¦È_ref = 180 - 0 = 180
¦È = 360 - ¦È_ref = 360 - 0 = 360
Therefore, the exact solutions for the equation sin^2 ¦È + 2cos ¦È = -2 in the range 0¡Ü¦È<360¡ã are ¦È = 180 and ¦È = 360.