I was asked to find maximum and minimum value of this $$2\sin^3x + \frac 34 \sin 2x + \frac 92 \cos 2x - 9\cos x - \frac 32 x + 6 $$ also $$0\leq x \leq \frac{\pi}{2}$$ and local extremum at $x=\frac{\pi}{3}$. The answer was maximum at $0$ and minimum at $\frac{\pi}{3}$. I used calculator but i found that at $x=\frac{\pi}{3}$ it yields -94.151 and when $x= \frac{\pi}{2}$ it yields -137.5
Is the answer sheet wrong?
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COMMENT.-Your calculations are wrong. You have correctly being $f$ the function $$f\left(\dfrac{\pi}{3}\right)\approx -0.37223916\\f\left(\dfrac{\pi}{2}\right)\approx 1.14380550\\f(0)=1.5$$
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Computing the derivative of $$f(x)=2\sin^3(x) + \frac 34 \sin (2x) + \frac 92 \cos (2x) - 9\cos (x) - \frac 32 x + 6$$ and using the tangent half-angle substitution, you wold end with $$f'(t)=\frac{6 t \left(3 t^2-1\right) (3 t^2-2t+3)}{\left(1+t^2\right)^3}$$ and notice that $\frac{6 (3 t^2-2t+3)}{\left(1+t^2\right)^3} >0 \quad \forall t$
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Let $$f(x)=2\sin^3x + \frac 34 \sin 2x + \frac 92 \cos 2x - 9\cos x - \frac 32 x + 6 .$$ Thus, $$f'(x)=6\sin^2x\cos{x}+\frac{3}{2}\cos2x-9\sin2x+9\sin{x}-\frac{3}{2}=$$ $$=6\cos{x}-6\cos^3x+3\cos^2x-3-18\sin{x}\cos{x}+9\sin{x}=$$ $$=3(2\cos{x}-1)-3\cos^2x(2\cos{x}-1)-9\sin{x}(2\cos{x}-1)=$$ $$=3(2\cos{x}-1)(\sin^2x-3\sin{x})=3(1-2\cos{x})(3-\sin{x})\sin{x}.$$ Can you end it now?
WolframAlpha confirms that the minimum is at $x=\frac\pi3$.
We can see from the graph in that link that $x=0$ is the maximum as this is where the turning point is (although WolframAlpha struggles to identify this)