Find all solutions of the question.
8 sin x - 9sqrt(2) = 6 sin x - 8sqrt(2)
This is not Calculus question.
8 sin x - 9 √2 = 6 sin x - 8 √2
Subtract 6 sin x to both sides
2 sin x - 9 √2 = - 8 √2
Add 9 √2 to both sides
2 sin x = √2
sin x = √2 / 2
On interval 0 to 2 π sin x = √2 / 2 for x = 45° = π / 4 rad
and
x = 135° = 3 / 4 π rad
Period of sin x is 2 π rad so the solutions are:
x = π / 4 + 2 π n
and
x = 3 / 4 π + 2 π n
where
n = ± 1 , ± 2 , ± 3...
By the way √2 / 2 is the same as 1 / √2 becouse:
√2 / 2 = √2 / √2 ∙ √2 = 1 / √2
My typo.
n = 0 , ± 1 , ± 2 , ± 3...
To solve the equation 8sin(x) - 9√2 = 6sin(x) - 8√2, we can follow these steps:
Step 1: Combine like terms
We have 8sin(x) - 9√2 = 6sin(x) - 8√2. Arrange the equation by gathering terms with sin(x) on one side and the constant terms on the other side:
8sin(x) - 6sin(x) = -8√2 + 9√2
Simplifying the equation gives us:
2sin(x) = √2
Step 2: Solve for sin(x)
Divide both sides of the equation by 2:
sin(x) = (√2) / 2
Step 3: Determine the solutions
To find the solutions of sin(x) = (√2) / 2, we look for angles x that have a sine value equal to (√2) / 2.
The sine function has a value of (√2) / 2 at two angles in the range [0, 2π]:
- π/4 (45 degrees) and 3π/4 (135 degrees)
So, the solutions to the equation 8sin(x) - 9√2 = 6sin(x) - 8√2 are:
x = π/4 + 2kπ, where k is an integer
x = 3π/4 + 2kπ, where k is an integer