Find the inverse of:
f(x) = -5 +11arccos(7^x - 9)
Find exactly all the real solutions for each of the following equations.
a. Sin(x) = - square root 3/2
b. 1 + cos(3x-pi/4) = 1/2
x = -5 + 11arccos(7^y-9)
(x+5)/11 = arccos(7^y-9)
7^y-9 = cos (x+5)/11
7^y = 9 + cos (x+5)/11
y = log7 (9 + cos (x+5)/11)
you know that sin π/3 = √3/2
so, 2kπ + π/3 or (2k+1)π - π/3
or (2k+1)(π/2)±(π/6)
1+cos(3x-π/4) = 1/2
cos(3x-π/4) = -1/2
3x-π/4 = 2π/3 or 4π/3
3x = 11π/12 or 19π/12 (+2π,4π,6π)
x = 11π/36 or 19π/36 (+2π/3,4π/3,2π)
To find the inverse of a function, we will follow these steps for the given function f(x) = -5 + 11arccos(7^x - 9):
Step 1: Replace f(x) with y.
y = -5 + 11arccos(7^x - 9)
Step 2: Swap the x and y variables.
x = -5 + 11arccos(7^y - 9)
Step 3: Solve for y.
x + 5 = 11arccos(7^y - 9)
Step 4: Isolate the arccos term.
(x + 5)/11 = arccos(7^y - 9)
Step 5: Apply the inverse cosine function to both sides.
cos((x + 5)/11) = 7^y - 9
Step 6: Isolate the exponential term.
cos((x + 5)/11) + 9 = 7^y
Step 7: Take the logarithm base 7 of both sides.
log7(cos((x + 5)/11) + 9) = y
Therefore, the inverse function is:
f^(-1)(x) = log7(cos((x + 5)/11) + 9)
Now, let's move on to find the real solutions for each of the following equations:
a. Sin(x) = -√3/2
To find the real solutions for this equation, we take the inverse sine (arcsin) of both sides since sine is an odd function. The inverse sine function gives us the angle whose sine value is equal to the given value.
arcsin(sin(x)) = arcsin(-√3/2)
x = π - π/3
Therefore, the real solution for this equation is:
x = 2π/3
b. 1 + cos(3x - π/4) = 1/2
To find the real solutions for this equation, we will manipulate it step by step:
1 + cos(3x - π/4) = 1/2
Subtract 1 from both sides:
cos(3x - π/4) = 1/2 - 1
cos(3x - π/4) = -1/2
Now, let's solve for 3x - π/4:
3x - π/4 = π ± π/3
3x = π/4 + π ± π/3
3x = (4π + 3π ± 4π)/12
3x = 21π/12 or 11π/12
Simplify it further:
x = 7π/12 or 11π/36
Therefore, the real solutions for this equation are:
x = 7π/12 or 11π/36