I have two questions as I review for a test tommorow, any help is much appreciated!!
Solve:
1) 4sin^2x=2cosx+1
2) cscx+cotx=1
Of course! I'd be happy to help you with your questions. Let's start with the first one:
1) To solve the equation 4sin^2x = 2cosx + 1, we can rewrite sin^2x as 1 - cos^2x using the Pythagorean identity: sin^2x + cos^2x = 1.
Substituting this expression into the equation, we have:
4(1 - cos^2x) = 2cosx + 1
Expanding the left side of the equation gives us:
4 - 4cos^2x = 2cosx + 1
Rearranging the terms to bring them to one side of the equation:
4cos^2x + 2cosx - 3 = 0
Now, let's solve this quadratic equation. We can use factoring or the quadratic formula to find the values of cosx that satisfy this equation. Factoring might seem a bit challenging, so we'll use the quadratic formula:
The quadratic formula states that given an equation ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation 4cos^2x + 2cosx - 3 = 0, a = 4, b = 2, and c = -3.
Applying the quadratic formula, we get:
cosx = (-2 ± √(2^2 - 4(4)(-3))) / (2(4))
Simplifying further:
cosx = (-2 ± √(4 + 48)) / 8
cosx = (-2 ± √52) / 8
cosx = (-2 ± 2√13) / 8
Breaking it down:
cosx = (-1 ± √13) / 4
Therefore, the solutions for cosx are:
cosx = (-1 + √13) / 4
cosx = (-1 - √13) / 4
Next, let's move on to the second question:
2) To solve the equation cscx + cotx = 1, we need to simplify and bring the terms to a common denominator.
First, let's rewrite cscx as 1/sinx and cotx as cosx/sinx, since these are the reciprocal identities.
Our equation becomes:
1/sinx + cosx/sinx = 1
Combining the fraction terms, we get:
(1 + cosx) / sinx = 1
To eliminate the fraction, we can multiply both sides of the equation by sinx:
1 + cosx = sinx
Rearranging the terms, we have:
1 = sinx - cosx
Now, we can square both sides of the equation to eliminate the square root:
1^2 = (sinx - cosx)^2
This simplifies to:
1 = sin^2x - 2sinxcosx + cos^2x
Using the Pythagorean identity sin^2x + cos^2x = 1, we can substitute this identity into the equation:
1 = 1 - 2sinxcosx + cos^2x
Rearranging the terms:
2sinxcosx - cos^2x = 0
Factoring out a common term:
cosx(2sinx - cosx) = 0
From this equation, we can see that either cosx = 0 or 2sinx - cosx = 0.
For cosx = 0, we know that cosx = 0 when x is equal to either π/2 or 3π/2, since those are the points where cosx equals zero.
For 2sinx - cosx = 0, we can rearrange the equation to isolate sinx:
2sinx = cosx
sinx = cosx/2
Using the Pythagorean identity again, sin^2x + cos^2x = 1, we can substitute:
sin^2x + (cosx/2)^2 = 1
Simplifying the equation:
sin^2x + cos^2x/4 = 1
(4sin^2x + cos^2x)/4 = 1
(5sin^2x + cos^2x)/4 = 1
5sin^2x + cos^2x = 4
Now we have a similar equation to the first one we solved.
From here, you can use the same steps we used for the first question to solve for the values of sinx and cosx that satisfy 5sin^2x + cos^2x = 4.
I hope this helps you understand how to solve these equations! Let me know if you have any further questions or need any additional clarification. Good luck on your test!