Find all solutions to 2 sin(theta) = sqrt(2) on the interval 0 less than or equal to theta less than 2 pi
Give your answers as exact values, as a list separated by commas.
2 sinØ = √2
sinØ = √2/2
from the 45-45-90 right-angled triangle you should recognize Ø = 45°
but the sine is positive in the I and II quads
so Ø = 45° or 135°
in radians , Ø = π/4 or 3π/4
thanks:)
To find all the solutions to the equation 2 sin(theta) = sqrt(2) on the interval 0 ≤ theta < 2π, we can follow these steps:
Step 1: Divide both sides of the equation by 2:
sin(theta) = sqrt(2)/2
Step 2: Determine the reference angle by finding the angle whose sine is equal to sqrt(2)/2. In this case, the reference angle is π/4, since sin(π/4) = sqrt(2)/2.
Step 3: Identify the quadrants where sine is positive or negative. In the first and second quadrants, sine is positive, while in the third and fourth quadrants, sine is negative.
Step 4: Express all possible theta values on the interval 0 ≤ theta < 2π using the reference angle and the quadrant information.
In the first and second quadrants, theta = π/4.
In the second and third quadrants, theta = π - π/4 = 3π/4.
Therefore, the solutions to the equation 2 sin(theta) = sqrt(2) on the interval 0 ≤ theta < 2π are:
θ = π/4, 3π/4
So, the exact values of the solutions are π/4 and 3π/4.
To find all solutions to the equation 2 sin(theta) = sqrt(2), we can follow these steps:
Step 1: Divide both sides of the equation by 2 to isolate sin(theta). This gives us sin(theta) = sqrt(2)/2.
Step 2: Recall the special values of sin(theta) for common angles. At angles 0°, 30°, 45°, 60°, and 90°, the value of sin(theta) is respectively 0, 1/2, sqrt(2)/2, sqrt(3)/2, and 1. From this, we see that sin(theta) = sqrt(2)/2 at angles 45° and 135° in the first and second quadrants.
Step 3: Use the symmetries of the sine function to find the other solutions. Since sin(theta) has a period of 2π, we can find solutions by adding integer multiples of 2π to the angles we found in Step 2. So, the other solutions occur at theta = 45° + 2πn and theta = 135° + 2πn, where n is an integer.
Step 4: Check if the solutions lie within the given interval 0 ≤ theta < 2π. The angles we found, 45°, 135°, 45° + 2π, and 135° + 2π, all fall within this interval.
Finally, we can write the solutions as a list separated by commas. The solutions are:
theta = 45°, 135°, 45° + 2π, 135° + 2π.