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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