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There's some way to find $x$ here ?

$$\sin(x)+\cos(x)-\tan(x)=0.4$$

Joel
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5 Answers5

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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

lab bhattacharjee
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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.)

Malper
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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.

user44197
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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.

Fly by Night
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  • Attention! Squaring operation introduce foreign solutions of the initial equation. – medicu Dec 25 '13 at 02:30
  • @medicu Please read my post carefully. I clearly state that all solutions to the original problem are solutions of the final expressions but that the converse need not be true, i.e. not all solutions of the final expressions need be solutions of the original problem. – Fly by Night Dec 25 '13 at 10:58
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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.

user2369284
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