Solve sinx+cosx=0
Thanks!
sinx=-cosx
tanx=-1
x= arctan(-1)
We're not supposed to find it using arctan. We have to get one side to be 2 factors that equal zero when multiplied.
To solve the equation sin(x) + cos(x) = 0, we can rewrite it in terms of a single trigonometric function.
One way to do this is by using the identity sin^2(x) + cos^2(x) = 1. Rearranging the terms, we get cos^2(x) = 1 - sin^2(x).
Substituting this into the original equation, we have sin(x) + (1 - sin^2(x)) = 0.
Expanding the equation, we get sin(x) + 1 - sin^2(x) = 0.
Rearranging the terms, we have -sin^2(x) + sin(x) + 1 = 0.
Now, we can treat this equation as a quadratic equation by replacing sin(x) with a variable, such as y.
So, we have -y^2 + y + 1 = 0.
To solve this quadratic equation, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = -1, b = 1, and c = 1. Substituting these values into the quadratic formula, we get:
y = (-1 ± √(1 - 4(-1)(1))) / (2(-1)).
Simplifying further, we have:
y = (-1 ± √(1 + 4)) / (-2).
y = (-1 ± √5) / (-2).
Now, to find the values of x, we substitute y back into the original equation sin(x) + (1 - sin^2(x)) = 0.
For y = (-1 + √5) / (-2):
sin(x) + (1 - sin^2(x)) = 0.
sin(x) + 1 - sin^2(x) = 0.
Substituting y = sin(x), we have:
y + 1 - y^2 = 0.
This can be rearranged to form a quadratic equation:
-y^2 + y + 1 = 0.
We have already solved this equation above and found the value of y.
Similarly, for y = (-1 - √5) / (-2), we have:
sin(x) + (1 - sin^2(x)) = 0.
Substituting y = sin(x), we have:
y + 1 - y^2 = 0.
This can be rearranged to form a quadratic equation:
-y^2 + y + 1 = 0.
Again, we have already solved this equation and found the value of y.
Therefore, the solutions to sin(x) + cos(x) = 0 are the values of x that correspond to the values of y we found from the quadratic equation: (-1 ± √5) / (-2).