Trigonometry equation $\sin(x)+\cos(x)-\tan(x)=0.4$ There's some way to find $x$ here ?
$$\sin(x)+\cos(x)-\tan(x)=0.4$$

This is a difficult problem to solve. If you write $\tan x$ as the quotient of $\sin x$ and $\cos x$ you get
$$\sin x + \cos x + \frac{\sin x}{\cos x} = \frac{2}{5}$$
It is possible to rearrange this formula, to get $\sin x$ as the subject:
$$\sin x = \frac{(2-5\cos x)\cos x}{5+5\cos x}$$
Squaring both sides, and using the fact that $\cos^2x+\sin^2x \equiv 1$, gives:
$$1-\cos^2x = \frac{(2-5\cos x)^2\cos^2 x}{25+25\cos^2 x}$$
We can cross multiply, and give ourselves a quartic equation to solve:
$$50\cos^4x + 30\cos^3x + 4\cos^2x - 50\cos x - 25 = 0 $$
We can perform as similar procedure and get ourselves the quartic equation:
$$50\sin^4x-70\sin^3x-46\sin^2x+70\sin x+21=0$$
Each of these quartics, when counted with multiplicity, will have exactly four solutions. Each of the four solutions will give four equations, e.g. $\cos x = u_1$, $\cos x = u_2$, $\cos x = u_3$ and $\cos x = u_4$. Similarly, $\sin x = v_1$, $\sin x= v_2$, $\sin x = v_3$ and $\sin x = v_4$. In each case, the $x$ values that solve your original problem will solve some of the $\cos x = u_i$ and some of the $\sin x = v_j$. The method I used has given us "If $x$ solves your equation then $\cos x$ and $\sin x$ will solve the two resulting quartics." The converse need not be true.
