i don't know how to work these problems out so i'll just pick a random one out from my home work and if someone could explain it to me step by step that would be very helpful.
Solve the tragicomic expression for all answers between 0 degrees and 360 degrees
2cos^2x-3cosx=2 where x is theta
Sure, I'd be happy to help you solve this problem step by step.
To solve the given trigonometric expression, we need to isolate the variable (theta or x) and then find its values within the given interval, which is 0 to 360 degrees.
Let's begin by rewriting the equation in a more convenient form:
2cos^2x - 3cosx = 2
Now, let's focus on the term "cos^2x". This can be rewritten as "(cosx)^2". By substituting this back into the equation, we get:
2(cosx)^2 - 3cosx = 2
Now, let's move all the terms to one side of the equation:
2(cosx)^2 - 3cosx - 2 = 0
This is now a quadratic equation in terms of "(cosx)". To solve it, we can either factor or use the quadratic formula. Let's factor it:
(2cosx + 1)(cosx - 2) = 0
Now, to find the values of "(cosx)" that satisfy the equation, we set each factor equal to zero:
2cosx + 1 = 0 and cosx - 2 = 0
Solving the first equation, we have:
2cosx = -1
cosx = -1/2
To find the angle "x" (or theta) that corresponds to cosx = -1/2 within the given interval (0 to 360 degrees), we consult the unit circle or the cosine function graph. We see that there are two possible angles: 120 degrees and 240 degrees.
Now, let's solve the second equation:
cosx - 2 = 0
cosx = 2
Here, we notice that the value of cosine cannot exceed 1, so there are no solutions within the given interval for this equation.
Therefore, the solutions to the original equation 2cos^2x - 3cosx = 2, within the interval 0 to 360 degrees, are x = 120 degrees and x = 240 degrees.