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
Sure! I'd be happy to help you with those questions. Let's start with the first one:
1) 4sin^2x = 2cosx + 1
To solve this equation, we can make use of the Pythagorean identity: sin^2(x) + cos^2(x) = 1. We can rewrite the equation in terms of cosine:
4(1 - cos^2(x)) = 2cosx + 1
Now, distribute the 4 on the left side:
4 - 4cos^2(x) = 2cosx + 1
Rearrange the equation and set it equal to zero:
4cos^2(x) + 2cosx - 3 = 0
Now, this is a quadratic equation in terms of cos(x). To solve it, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
cos(x) = (-b ± √(b^2 - 4ac))/(2a)
For our equation, a = 4, b = 2, and c = -3. Substituting these values in, we get:
cos(x) = (-2 ± √(2^2 - 4(4)(-3))) / (2*4)
cos(x) = (-2 ± √(4 + 48)) / 8
cos(x) = (-2 ± √52) / 8
Simplifying further:
cos(x) = (-2 ± 2√13) / 8
cos(x) = (-1 ± √13) / 4
Now, we have two possible solutions for cos(x). To find the values of x, we take the inverse cosine (also known as arccos) of each solution:
x = arccos((-1 ± √13) / 4)
Now, we can use a calculator to find the value of x. Remember to check if your calculator is in degrees or radians mode, depending on the given instructions.
Now, let's move on to the second question:
2) cscx + cotx = 1
We can rewrite this equation in terms of sine and cosine using the reciprocal identities:
1/sinx + cosx/sinx = 1
To get rid of the denominators and have a single fraction, we need to find the common denominator, which is sinx. Then, we can combine the terms:
(1 + cosx) / sinx = 1
Cross-multiply the equation:
1 + cosx = sinx
Now, use the Pythagorean identity to rewrite sinx in terms of cosx:
1 + cosx = √(1 - cos^2(x))
Square both sides of the equation to eliminate the square root:
1 + 2cosx + cos^2(x) = 1 - cos^2(x)
Combine like terms:
2cosx + 2cos^2(x) = 0
Factor out the common factor:
2cosx(1 + cosx) = 0
Set each factor equal to zero and solve for x:
cosx = 0 or 1 + cosx = 0
x = arccos(0) or cosx = -1
Again, use a calculator to find the value of x from the arccos function. For the second equation, cosx = -1 has a solution of x = π (or 180 degrees, depending on the units given).
I hope this helps! Let me know if you have any further questions. Good luck on your test!