Hi everyone I would like to ask a thing about the following equation:
$$\cos(x) + \sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = 0$$
It is trigonometric and irrational, the root's index is 4 (even root).
Now, given that:
$$a = b \implies a^4 = b^4 $$
But:
$a^4 = b^4\,\,$ does not necessarily mean that $a = b$, how should I handle it with the equation above? Should I care about the conditions of existence of the 4th root?
I mean:
$1 - \frac{4}{3}\cos(2x) - \sin^4(x) \ge 0$ ?
Or, can I just do the following:
$$\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,cos(x)$$
And raise both members to the power of 4 so that:
$$1 - \frac{4}{3}\cos(2x) - \sin^4(x) = \cos^4(x)$$
And solve the equation remembering that $\cos(x)$ must be $\le 0$ ???
Is this enough? Will solving the system:
\begin{equation} \begin{cases} 1 - \frac{4}{3}\cos(2x) - \sin^4(x) = \cos^4(x)\\ \cos(x) \le 0 \end{cases} \end{equation}
give me all the solutions for the equation but only the true solutions for which $1 - \frac{4}{3}\cos(2x) - \sin^4(x)$ is greater than or equal to 0?
Or should I compute all the cases under the root when the radicand can be negative?
Just need some elucidation. Thanks for the attention!
Your start is correct.
All the solutions of the equation:
$$1 - \frac{4}{3}\cos(2x) - \sin^4(x) = \cos^4(x)$$
necessarily have $1 - \frac{4}{3}\cos(2x) - \sin^4(x)\geq 0$. There is no reason to add that condition.
Adding the condition that you need $\cos(x)\leq 0$ gives you what you want.
The next step is to write everything as an equation of $\cos^2(x)$ - both $\cos(2x)$ and $\sin^4(x)$ can be written in terms of $\cos^2(x)$.
If $X=\cos^2(x)$, you get the equation:
$$1-\frac{4}{3}(2X-1) -(1-X)^2 = X^2$$
Solve for $X$, choose only the non-negative roots, then find solve $\cos(x)=-\sqrt{X}$ for those $X$.