Today, in our class, we received a trigonometric equation

[Bobs]

Today,in our class, we received a trigonometric equation
$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$

Here is my attempt:

 

[.Scott]

<Moderator's note: Solution removed.>

Here's what it looks like graphed out:

 

[.Scott]

I cannot read a lot of your handwriting, but from what I can read, I suspect you do not have it.
I am assuming you are only looking for the real (non complex) roots.

 

[Ray Vickson]

Today,in our class, we received a trigonometric equation
$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$

Here is my attempt:
View attachment 227423

Here is one way that will work, but be prepared for pages and pages of algebra---or better yet, use a computer algebra package, as I did: I used Maple.

You can use Euler's formulas (see https://en.wikipedia.org/wiki/Euler's_formula) to write
$$\cos(x) = \frac{1}{2} (u + v), \;\; \sin(x) = \frac{1}{2i} (u - v), $$
where
$$u = e^{ix}, \;\; v = e^{-ix} = \frac 1 u.$$
Also, $\cos(2x) = (1/2)(e^{2ix} + e^{-2ix}) = (1/2)(u^2 + v^2).$ That gives your function
$$f(x) = \sin^{10} (x) + \cos^{10} (x) - \frac{29}{16} \cos^4 (2x)$$
as
$$f = -\left( \frac u 2 - \frac v 2 \right)^{10} + \left( \frac u 2 + \frac v 2 \right)^{10}
- \frac{29}{16} \left( \frac{u^2}{2} + \frac{v^2}{2} \right)^4.$$
Expand it all out, set $v = 1/u$, then simplify. You will get an expression of the form
$$f = -\frac{P(u^4)}{32 u^8}, $$
where $P(y)$ is a 4th degree polynomial in $y$ with positive integer coefficients. Furthermore, $P(y)$ factors into two quadratics with integer coefficients, so the equation $f(x) = 0$ translates into the easily solvable equation $P(y) = 0$, where $y = u^4 = e^{4ix}.$

 

[ehild]

Today,in our class, we received a trigonometric equation
$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$

Here is my attempt:
View attachment 227423

I can not read the picture. What is your solution?
sin¹⁰(x) + cos¹⁰(x) can be written as (sin²(x))⁵+(cos²(x))⁵,
Using the identities $\sin^2(x)=\frac{1-\cos(2x)}{2}$ and $\cos^2(x)=\frac{1+\cos(2x)}{2}$. the equation has the variable y=cos(2x). Expanding and simplifying, you get a quadratic equation for y², with very simple solution.