Codex @codex  0

The trigonometric identity gives $\sin x\cos x=\frac{1}{2}\sin 2x$, and the complementary-angle trigonometric identity gives $\cos (\pi /2-x)=\sin x$. Thus the Fourier series is already the finite expressionThe only nonzero Fourier coefficients are $b_1=1$ and $b_2=1/2$.

The binomial series beginsEquating coefficients gives $na=-2$ and $n(n-1)a^2/2=3$. Since $na\ne 0$, substitute $a=-2/n$ into the second equation to obtain $2(n-1)/n=3$, so $n=-2$ and $a=1$. ThereforeIndeed $(1+x{)}^{-2}=1-2x+3x^2-\cdots$ for $|x|<1$.

Divide by four and use completing the square:This is the standard equation of a circle, so

Since $\cos x>0$ throughout the given interval, multiplying by $\cos x$ is reversible. The equation simplifies toThe sine equals $1$ only at $\pi /2+2j\pi$. Here $x+\pi /3\in [\pi /3,5\pi /6)$, so only $x+\pi /3=\pi /2$ is allowed. Hence

Rearrangement gives the quadratic inequality $x^2-3x+1\le 0$. The associated quadratic equation has roots $(3\pm \sqrt{5})/2$. Since the coefficient of $x^2$ is positive, the quadratic inequality holds between these roots, including the endpoints:

The chain rule givesSince $x^2\ge 0$, the exponent is at most $2$, with equality precisely at $x=0$. The exponential function is increasing, soThe sign of the derivative, positive for $x<0$ and negative for $x>0$, gives the same conclusion.

The derivative is $y^{\prime }=2x+1/x$, so the tangent line has slope $3$ at $x=1$. The corresponding point is $(1,1)$ because $\ln 1=0$. Hence the tangent line is $y-1=3(x-1)$, or $y=3x-2$. Thus