Answer:
Explanation:
cos(x+50) = 1/2
We know that cos(60) = 1/2, so we can write:
cos(x+50) = cos(60)
Using the identity cos(a) = cos(b) if and only if a = ±b + 2πn, we get:
x+50 = ±60 + 2πn
Solving for x, we have:
x = -50 ±60 + 2πn
x = 10 + 2πn or x = -110 + 2πn
Since we want to find all values of x between 0 and 180, we only need to consider the values of x that satisfy 0 ≤ x ≤ 180.
For x = 10 + 2πn, we have:
0 ≤ 10 + 2πn ≤ 180
-10/2π ≤ n ≤ 85/2π
For x = -110 + 2πn, we have:
0 ≤ -110 + 2πn ≤ 180
-35/2π ≤ n ≤ 25/2π
Therefore, the solutions for cos(x+50) = 1/2 in the interval [0, 180] are:
x = 10° + 360°n or x = 150° + 360°n, where n is an integer.
sin(2x) = -0.6
We know that sin(θ) = -0.6 has two solutions in the interval [0, 360]: θ ≈ -36.87° and θ ≈ 216.87° (using a calculator).
Using the double angle identity sin(2x) = 2sin(x)cos(x), we can write:
2sin(x)cos(x) = -0.6
Dividing both sides by 2cos(x), we get:
sin(x) = -0.3/cos(x)
Using the identity cos²(x) + sin²(x) = 1, we can substitute sin²(x) = 0.09/cos²(x) and simplify to get:
cos³(x) - 3cos(x) + 0.9 = 0
We can solve this equation using numerical methods or approximations. One possible approximation is to use the intermediate value theorem and test for sign changes in the function f(x) = cos³(x) - 3cos(x) + 0.9:
f(0) = 0.9 > 0
f(π/2) ≈ -0.84 < 0
f(π) ≈ 0.49 > 0
Therefore, there is a root of f(x) = 0 in the interval (0, π/2) and another root in the interval (π/2, π).
Using a numerical solver or more advanced methods, we can find the approximate values of x that satisfy cos³(x) - 3cos(x) + 0.9 = 0 in the intervals (0, π/2) and (π/2, π):
x ≈ 68.58° or x ≈ 111.42°
Therefore, the solutions for sin(2x) = -0.6 in the interval [0, 180] are:
x ≈ 34.29° or x ≈ 55.71°