Avoid squaring wherever practicable as it immediately introduces extraneous roots

Using Weierstrass substitution we have

$$frac2t1+t^2+frac1-t^21+t^2-frac2t1-t^2=0.4$$ where $t= anfrac x2$

On rearrangement we have a Biquadratic Equation in $t$ with no extraneous root, find the answer here


Let $y=cos(x)$. Then the equation becomes$$pmsin(cos^-1(y))+ymp an(cos^-1(y))=0.4$$$$pmsqrt1-y^2+ympfracsqrt1-y^2y=0.4$$$$2 y^4-2.8 y^3+0.16y^2+2y-1=0$$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 $x$ ($x approx 2.54982$ & $x approx 0.792437$), & all the solutions are obtained from these by adding integer multiples of $2pi$.

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(If you wanted to, you could use the formula for roots of a quartic khổng lồ get an analytic solution, but due lớn messiness I"ve refrained from doing so here.)


For a numerical solution, Newton"s method is the fall back option:

Let $$f(x) = sin(x)+cos(x)- an(x) -0.4$$Then$$f"(x) = cos(x)-sin(x) -sec^2(x) $$Newton"s iterative formula is $$x_n+1 = x_n - fracf(x_n)f"(x_n)$$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 $pi/4$ và the other close to lớn 2.5. Starting with these two initial guesses one obtains$$eginalignx &=0.79243740562806 && hboxsolution close khổng lồ $pi/4$ \x &=2.549822120435232&& hboxsolution close lớn 2.5 \endalign$$

I can"t think of any trigonometric identities that will give these two answers.


This is a difficult problem khổng lồ solve. If you write $ an x$ as the quotient of $sin x$ & $cos x$ you get$$sin x + cos x + fracsin xcos x = frac25$$It is possible to rearrange this formula, to lớn get $sin x$ as the subject:$$sin x = frac(2-5cos x)cos x5+5cos x$$Squaring both sides, and using the fact that $cos^2x+sin^2x equiv 1$, gives:$$1-cos^2x = frac(2-5cos x)^2cos^2 x25+25cos^2 x$$We can cross multiply, & give ourselves a quartic equation to solve:$$50cos^4x + 30cos^3x + 4cos^2x - 50cos x - 25 = 0 $$We can perform as similar procedure and get ourselves the quartic equation:$$50sin^4x-70sin^3x-46sin^2x+70sin x+21=0$$Each of these quartics, when counted with multiplicity, will have exactly four solutions. Each of the four solutions will give four equations, e.g. $cos x = u_1$, $cos x = u_2$, $cos x = u_3$ và $cos x = u_4$. 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 $cos x = u_i$ & some of the $sin x = v_j$. The method I used has given us "If $x$ solves your equation then $cos x$ & $sin x$ will solve the two resulting quartics.

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" The converse need not be true.