Question

Solve the following equation and state the general solution for all values of x in exact form. Show all steps of your algebraic solution.

𝑠𝑖𝑛^2(𝑥) − √2𝑐𝑜𝑠(𝑥) = 𝑐𝑜𝑠^2(𝑥) + √2𝑐𝑜𝑠(𝑥) + 2

I find this question to be really difficult. Could anyone help me?

Answer 1

change the sin^2 x to 1 - cos^2 x and you have only cos x in your equation:

1 - cos^2 x - √2cosx = cos^2 x + √2cosx + 2
2cos^2 x + 2√2cosx + 1 = 0
This factors to
(√2 cosx + 1)^2 = 0
cosx = -1/√2
we know the cosine is negative in quads II and III
x = 135° or x = 225°
or in radians, x = 3π/4, x = 5π/4

The period of cosx is 2π
so we have:
x = 3π/4 + 2kπ , 5π/4 + 2kπ, where k is an integer.

Answer 2

since sin^2x = 1-cos^2x, let u=cosx and

make things easier to read and write it as
1 - u^2 - √2 u = u^2 + √2 u + 2
2u^2 + 2√2 u + 1 = 0
(√2 u + 1)^2 = 0
u = -1/√2
so now we have
cosx = -1/√2
x = π ± π/4 + 2kπ

Answer 3

[AI-generated answer — Step-by-Step Bot]

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) - √2cos(x) = cos^2(x) + √2cos(x) + 2

Step 2: Move all terms to one side of the equation:
sin^2(x) - cos^2(x) - √2cos(x) - √2cos(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) - 2√2cos(x) - 2 = 0

Step 4: Rearrange the equation:
cos(2x) + 2√2cos(x) + 2 = 0

Step 5: Let's make a substitution to simplify the equation. Let z = cos(x). Now we have:
cos(2x) + 2√2z + 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 + 2√2z + 2 = 0

Step 7: Simplify:
2cos^2(x) + 2√2z + 1 = 0

Step 8: Multiply the entire equation by 2:
4cos^2(x) + 4√2z + 2 = 0

Step 9: Use quadratic formula to solve for z:
z = (-b ± √(b^2 - 4ac))/2a

Here, a = 4, b = 4√2, and c = 2.

Substituting the values into the quadratic formula:
z = (-4√2 ± √((4√2)^2 - 4(4)(2)))/(2(4))

Step 10: Simplify further:
z = (-4√2 ± √(32 - 32))/(8)

Since the square root of (32 - 32) is zero, we can simplify the equation to:

z = (-4√2 ± 0)/(8)

Step 11: Simplify even further:
z = -4√2/8

Simplifying the fraction:
z = -√2/2

Step 12: Recall that z = cos(x):
cos(x) = -√2/2

Step 13: We know that cos(x) = -√2/2 at specific angles. These angles are π/4 and 7π/4. Therefore, we have two equations:

x = π/4 and x = 7π/4

These are the general solutions for the equation sin^2(x) − √2cos(x) = cos^2(x) + √2cos(x) + 2.

Answer 4

[AI-generated answer — Explain Bot]

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) - √2cos(x) = cos^2(x) + √2cos(x) + 2

Notice that we have a common term, √2cos(x), on both sides of the equation. Let's isolate the terms with √2cos(x) on one side by subtracting √2cos(x) from both sides:

sin^2(x) - cos^2(x) - 2√2cos(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) - 2√2cos(x) = 2

Next, let's multiply both sides of the equation by -1 to get rid of the negative sign:

cos(2x) + 2√2cos(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) + 2√2) = -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) + 2√2 = 0, then the left-hand side of the equation becomes 0, and we get:
cos(x) = -2√2

Taking the cosine inverse of both sides, we have:

x = arccos(-2√2)

Now, we can use a calculator to approximate the value of arccos(-2√2).

The general solution is to add a multiple of 2π to the value we obtained, since cosine is a periodic function with a period of 2π.

So, the general solution for all values of x in exact form is:

x = arccos(-2√2) + 2πn

where n is an integer.