Trigonometry equation $\sin(x)+\cos(x)-\tan(x)=0.4$ There's some way to find $x$ here ?
$$\sin(x)+\cos(x)-\tan(x)=0.4$$
 A: Avoid squaring wherever practicable as it immediately introduces extraneous roots 
Using  Weierstrass substitution we have
$$\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-\frac{2t}{1-t^2}=0.4$$ where $t=\tan\frac x2$
On rearrangement we have a Biquadratic Equation in $t$ with no extraneous root, find the answer here
A: Let $y=\cos(x)$. Then the equation becomes
$$\pm\sin(\cos^{-1}(y))+y\mp\tan(\cos^{-1}(y))=0.4$$
$$\pm\sqrt{1-y^2}+y\mp\frac{\sqrt{1-y^2}}{y}=0.4$$
$$2 y^4-2.8 y^3+0.16y^2+2y-1=0$$
You can solve this using the usual methods for solving a quartic (yielding real solutions $y \approx -0.829954$ and $y \approx 0.702112$). Using $x = \cos^{-1}(y)$ gives particular solutions for $x$ ($x \approx 2.54982$ and $x \approx 0.792437$), and all the solutions are obtained from these by adding integer multiples of $2\pi$.
(If you wanted to, you could use the formula for roots of a quartic to get an analytic solution, but due to messiness I've refrained from doing so here.)
A: For a numerical solution, Newton's method is the fall back option:
Let $$f(x) = \sin(x)+\cos(x)-\tan(x) -0.4$$
Then
$$f'(x) = \cos(x)-\sin(x) -\sec^2(x) $$
Newton's iterative formula is 
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
and we start with $x_0$ some good initial guess (this is the tricky part!)
Rough sketch of the function shows that the two roots, one close to $\pi/4$ and the other close to 2.5. Starting with these two initial guesses one obtains
$$
\begin{align}
x &=0.79243740562806 && \hbox{solution close to $pi/4$} \\
x &=2.549822120435232&& \hbox{solution close to 2.5} \\
\end{align}
$$
I can't think of any trigonometric identities that will give these two answers.
A: 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.
A: Convert $\tan x = \frac{\sin x}{\cos x}$
Now we can write $\cos x$ as $\sqrt{1-\sin^2x}$ 
Now you will get an equation in sin x which you can solve.
