Solve the given triginometric equation analytically. Use values of of x for 0 less than or equal to x less than 2pi
sin^2x - 2sinx = 1
How do I get towards the answer of 3.569, 5.856
Thank you again
hint:
sin^2 x - 2sinx - 1 = 0
use the quadratic formula to solve, your answer will be in radians.
I got sinx = 1-√2 or sinx = 1+√2 (too big, sine has to be ≤ 1)
from sinx = 1-√2 = -.4142 x must be in quad III or IV
related acute angle in radians is .427
in III x = .427 + pi = 3.569
in IV x = 2pi - .427 = 5.856
can you take it from here?
To solve the given trigonometric equation, let's start by rewriting it:
sin^2(x) - 2sin(x) = 1
Notice that we have a quadratic equation in terms of sin(x). We can use a substitution to simplify this equation further. Let's substitute sin(x) with a new variable, say u:
Let u = sin(x)
Now, our equation becomes:
u^2 - 2u = 1
Rearranging the equation, we have:
u^2 - 2u - 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -2, and c = -1. Plugging in these values, we have:
u = (2 ± √((-2)^2 - 4(1)(-1))) / (2(1))
Simplifying further:
u = (2 ± √(4 + 4)) / 2
u = (2 ± √(8)) / 2
u = (2 ± 2√2) / 2
u = 1 ± √2
So we have two possible solutions for u:
u = 1 + √2 or u = 1 - √2
Now we need to solve for x using the values of u.
Let's consider the first solution, u = 1 + √2:
sin(x) = 1 + √2
To find the values of x, we take the inverse sine (or arcsin) of both sides:
x = arcsin(1 + √2)
To find the value of x, we use a calculator or a trigonometric table and take the inverse sine of 1 + √2. In this case, the calculator gives us 3.5688 (approximately).
Now let's consider the second solution, u = 1 - √2:
sin(x) = 1 - √2
Again, we take the inverse sine of both sides:
x = arcsin(1 - √2)
Using a calculator or a trigonometric table, we find that the inverse sine of 1 - √2 is approximately 5.8562.
Therefore, the solutions to the given trigonometric equation in the interval 0 ≤ x < 2π are approximately x = 3.569 and x = 5.856.