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.