JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Trigonometric Ratios and Identities

Writing $3{\sin }^{2}x+12\cos x-3=-3{\cos }^{2}x+12\cos x$ and setting $c=\cos x\in [-1,1]$, the function $h(c)=-3c^2+12c$ has its vertex at $c=2$, so it increases throughout $[-1,1]$, running from $h(-1)=-15$ to $h(1)=9$. Thus the equation has a solution exactly when $-15\le p\le 9$. The integers from $-15$ to $9$ sum to $(1+2+\cdots +9)-(1+2+\cdots +15)=45-120=-75$. The sum of all integral values of $p$ is $-75$.