Solve the following equation and state the general solution for all values of x in exact form. Show all steps of your algebraic solution.
š šš^2(š„) ā ā2ššš (š„) = ššš ^2(š„) + ā2ššš (š„) + 2
I find this question to be really difficult. Could anyone help me?
change the sin^2 x to 1 - cos^2 x and you have only cos x in your equation:
1 - cos^2 x - ā2cosx = cos^2 x + ā2cosx + 2
2cos^2 x + 2ā2cosx + 1 = 0
This factors to
(ā2 cosx + 1)^2 = 0
cosx = -1/ā2
we know the cosine is negative in quads II and III
x = 135Ā° or x = 225Ā°
or in radians, x = 3Ļ/4, x = 5Ļ/4
The period of cosx is 2Ļ
so we have:
x = 3Ļ/4 + 2kĻ , 5Ļ/4 + 2kĻ, where k is an integer.
since sin^2x = 1-cos^2x, let u=cosx and
make things easier to read and write it as
1 - u^2 - ā2 u = u^2 + ā2 u + 2
2u^2 + 2ā2 u + 1 = 0
(ā2 u + 1)^2 = 0
u = -1/ā2
so now we have
cosx = -1/ā2
x = Ļ Ā± Ļ/4 + 2kĻ
Sure, I can help you with that. Let's solve the equation step by step.
Step 1: Simplify the equation by combining like terms:
sin^2(x) - ā2cos(x) = cos^2(x) + ā2cos(x) + 2
Step 2: Move all terms to one side of the equation:
sin^2(x) - cos^2(x) - ā2cos(x) - ā2cos(x) - 2 = 0
Step 3: Recall the trigonometric identity: sin^2(x) - cos^2(x) = -cos(2x). Substituting this identity into the equation:
-cos(2x) - 2ā2cos(x) - 2 = 0
Step 4: Rearrange the equation:
cos(2x) + 2ā2cos(x) + 2 = 0
Step 5: Let's make a substitution to simplify the equation. Let z = cos(x). Now we have:
cos(2x) + 2ā2z + 2 = 0
Step 6: Using the double angle formula for cosine, we know that cos(2x) = 2cos^2(x) - 1. Substituting this into the equation:
2cos^2(x) - 1 + 2ā2z + 2 = 0
Step 7: Simplify:
2cos^2(x) + 2ā2z + 1 = 0
Step 8: Multiply the entire equation by 2:
4cos^2(x) + 4ā2z + 2 = 0
Step 9: Use quadratic formula to solve for z:
z = (-b Ā± ā(b^2 - 4ac))/2a
Here, a = 4, b = 4ā2, and c = 2.
Substituting the values into the quadratic formula:
z = (-4ā2 Ā± ā((4ā2)^2 - 4(4)(2)))/(2(4))
Step 10: Simplify further:
z = (-4ā2 Ā± ā(32 - 32))/(8)
Since the square root of (32 - 32) is zero, we can simplify the equation to:
z = (-4ā2 Ā± 0)/(8)
Step 11: Simplify even further:
z = -4ā2/8
Simplifying the fraction:
z = -ā2/2
Step 12: Recall that z = cos(x):
cos(x) = -ā2/2
Step 13: We know that cos(x) = -ā2/2 at specific angles. These angles are Ļ/4 and 7Ļ/4. Therefore, we have two equations:
x = Ļ/4 and x = 7Ļ/4
These are the general solutions for the equation sin^2(x) ā ā2cos(x) = cos^2(x) + ā2cos(x) + 2.
Of course! I'd be happy to help you solve this equation step by step.
Let's start by simplifying the equation:
sin^2(x) - ā2cos(x) = cos^2(x) + ā2cos(x) + 2
Notice that we have a common term, ā2cos(x), on both sides of the equation. Let's isolate the terms with ā2cos(x) on one side by subtracting ā2cos(x) from both sides:
sin^2(x) - cos^2(x) - 2ā2cos(x) = 2
Now, we can use a trigonometric identity to simplify the left-hand side of the equation. The identity we'll use is:
sin^2(x) - cos^2(x) = -cos(2x)
So, substituting this identity into the equation, we get:
-cos(2x) - 2ā2cos(x) = 2
Next, let's multiply both sides of the equation by -1 to get rid of the negative sign:
cos(2x) + 2ā2cos(x) = -2
Now, let's focus on the left-hand side of the equation. We can factor out a common term of cos(x):
cos(x)(cos(x) + 2ā2) = -2
Now, we have two possibilities:
1. If cos(x) = 0, then the left-hand side of the equation becomes 0, and we get:
0 = -2, which is not possible. So, cos(x) cannot be 0.
2. If cos(x) + 2ā2 = 0, then the left-hand side of the equation becomes 0, and we get:
cos(x) = -2ā2
Taking the cosine inverse of both sides, we have:
x = arccos(-2ā2)
Now, we can use a calculator to approximate the value of arccos(-2ā2).
The general solution is to add a multiple of 2Ļ to the value we obtained, since cosine is a periodic function with a period of 2Ļ.
So, the general solution for all values of x in exact form is:
x = arccos(-2ā2) + 2Ļn
where n is an integer.