Hello! I am a little stuck (probably because I cannot remember my trigonometric functions). Does anyone know how to solve this equation?
[-sin(x)]^2-2cosx
Hello! I can definitely help you with that equation. The first step is to expand the equation using trigonometric identities.
Let's start with the term [-sin(x)]^2. To expand it, you can square the negative sign as well as the sin(x) term, giving you sin^2(x).
Now, let's substitute this result back into the equation:
sin^2(x) - 2cos(x)
Next, we can recall the Pythagorean identity, which states that sin^2(x) + cos^2(x) = 1. Rearranging this equation, we get sin^2(x) = 1 - cos^2(x).
Substituting this back into the equation:
(1 - cos^2(x)) - 2cos(x)
Now, we have an equation solely involving the cosine function.
Let's simplify it further:
1 - cos^2(x) - 2cos(x)
Using the distributive property, we get:
- cos^2(x) - 2cos(x) + 1
Now, we have a quadratic equation in terms of the variable cos(x). To solve it, we can use factoring or the quadratic formula.
To factor, we look for two numbers that multiply to give -1 and add up to -2. These numbers are -1 and 1. So, we can rewrite the equation as:
(-cos(x) + 1)(cos(x) + 1)
Now, we have two possible solutions:
1) -cos(x) + 1 = 0
Solving for cos(x), we subtract 1 from both sides:
-cos(x) = -1
Multiplying both sides by -1:
cos(x) = 1
2) cos(x) + 1 = 0
Solving for cos(x), we subtract 1 from both sides:
cos(x) = -1
These are the possible solutions for the given equation. Remember that cos(x) represents the cosine function, so the solutions for x will depend on the range of values you are considering.