Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π).

8. cos2x=cosx
10. 2cos^2x+cosx=cos2x
Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your algebraic work.
24. cos2x+sinx=0

Have you tried the double angle formulas?

Yes. It is just too hard and it would be nice if you would just explain it to me or show me an example. Please.

Hard? Why would anyone want to do something easy?

10.2cos^2x+cosx=cos2x
2cos^2x+cosx=cosx= cos(x)cos(x) - sin(x)sin(x) = cos²(x) - sin²(x)
2 cos^2 x+cosx=cos^2x-(1-cos^2x)
2 cos^2 x+cosx=2cos^2x-1
cosx=-1
x=PI

the only way to make these easy is to do dozens of them, not just the assignments. The double angle formulas are critical to know.

24.
cos 2x + sin x=0
1 - 2sin^2 x + sin x = 0
rearrange things a bit to the usual order:

2sin^2 x - sin x - 1 = 0
(2sin x + 1)(sin x - 1) = 0

so, either sin x = -1/2 or sin x = 1

sin x = -1/2: x = 7pi/6 or 11pi/6
sin x = 1: x = pi/2

Thanks for your help you guys.

Sure! Let's solve each of these equations one by one:

8. cos2x = cosx

To solve this equation algebraically, we can use a trigonometric identity. The double-angle identity for cosine states that:

cos(2x) = 2cos^2(x) - 1

Substituting this into the equation, we have:

2cos^2(x) - 1 = cosx

Rearranging the equation, we get:

2cos^2(x) - cosx - 1 = 0

Now, let's solve this quadratic equation for cosx. We can factor it, use the quadratic formula, or complete the square. Factoring is the most convenient method here:

(2cosx + 1)(cosx - 1) = 0

Setting each factor equal to zero, we have two possibilities:

2cosx + 1 = 0 -> cosx = -1/2
cosx - 1 = 0 -> cosx = 1

Now, let's find the corresponding values of x within the interval [0,2π).

For cosx = -1/2, we need to find the values of x for which cosx = -1/2. These occur at π/3 and 5π/3.
For cosx = 1, we have x = 0 and 2π.

Thus, the solutions to the equation cos2x = cosx within the interval [0,2π) are x = {0, π/3, 5π/3, 2π}.

10. 2cos^2x + cosx = cos2x

Similarly to the previous equation, we can use the double-angle identity for cosine to rewrite this equation:

2cos^2(x) + cosx = 2cos^2(x) - 1

Rearranging the equation, we get:

cos^2(x) + cosx - 1 = 0

This equation involves a quadratic in terms of cosx. Let's solve it:

(cosx - 1)(cosx + 1) + cosx - 1 = 0

Simplifying further, we have:

cos^2(x) + 2cosx - 1 = 0

Now we can factor this equation:

(cosx - 1)(cosx + 1) + 2(cosx - 1) = 0

(cosx - 1)(cosx + 1 + 2) = 0

(cosx - 1)(cosx + 3) = 0

Setting each factor equal to zero, we obtain:

cosx - 1 = 0 -> cosx = 1
cosx + 3 = 0 -> cosx = -3 (extraneous solution)

Since cosx cannot be greater than 1 or less than -1, we discard -3 as an extraneous solution. Therefore, the only solution is cosx = 1. In the given interval [0,2π), this occurs at x = 0 and 2π.

Hence, the solution to the equation 2cos^2x + cosx = cos2x within the interval [0,2π) is x = {0, 2π}.

24. cos2x + sinx = 0

To solve this equation algebraically, we can use a trigonometric identity. The double-angle identity for cosine states that:

cos(2x) = 1 - 2sin^2(x)

Substituting this into the equation, we have:

1 - 2sin^2(x) + sinx = 0

Now we have a quadratic equation in terms of sinx. Let's rearrange it:

2sin^2(x) - sinx + 1 = 0

This equation can be solved using various methods: factoring, the quadratic formula, or completing the square. However, in this case, it does not factor or provide rational solutions. So, we will use a grapher to find the approximate solutions.

Using a graphing calculator or an online graphing software, plot the graph of the function y = 2sin^2(x) - sinx + 1. Look for the x-values (within the interval [0,2π)) where the graph intersects the x-axis, which indicates the solutions to the equation.

Remember, we only use the grapher to support our algebraic work and find approximate values. The exact solutions should be obtained algebraically.

So, the approximate solutions to the equation cos2x + sinx = 0 within the interval [0,2π) can be determined using a grapher, while the exact solutions can be found using algebraic methods.