Mathematics - Trigonometric Equation Question with Solution | TestHub

The inequality $\left(\sin x-\frac{1}{2}\right)\left(\cos x+\frac{1}{2}\right)>0$ holds if either both factors are positive or both are negative. Case 1:$\sin x-\frac{1}{2}>0$ and $\cos x+\frac{1}{2}>0$. This means $\sin x>\frac{1}{2}$ and $\cos x>-\frac{1}{2}$. In $[0^{\circ },360^{\circ }]$, $\sin x>\frac{1}{2}$ for $x\in (30^{\circ },150^{\circ })$. Also, $\cos x>-\frac{1}{2}$ for $x\in [0^{\circ },120^{\circ })\cup (240^{\circ },360^{\circ }]$. The intersection of these intervals is $x\in (30^{\circ },120^{\circ })$. Integer values of $x$ in this range are $31,32,\dots ,119$. The count is $119-31+1=89$. Case 2:$\sin x-\frac{1}{2}<0$ and $\cos x+\frac{1}{2}<0$. This means $\sin x<\frac{1}{2}$ and $\cos x<-\frac{1}{2}$. In $[0^{\circ },360^{\circ }]$, $\sin x<\frac{1}{2}$ for $x\in [0^{\circ },30^{\circ })\cup (150^{\circ },360^{\circ }]$. Also, $\cos x<-\frac{1}{2}$ for $x\in (120^{\circ },240^{\circ })$. The intersection of these intervals is $x\in (150^{\circ },240^{\circ })$. Integer values of $x$ in this range are $151,152,\dots ,239$. The count is $239-151+1=89$. The total number of integer values of$x$ is the sum of counts from Case 1 and Case 2, which is $89+89=178$.