Trigonometric RatiosmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Range Sum of Integer p for a Trig Equation = -75 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The sum of all the integral values of pp such that the equation 3sin⁡2x+12cos⁡x−3=p3\sin^2 x+12\cos x-3=p, x∈Rx\in\mathbb{R}, has at least one solution, is
A−54-54
B−60-60
C−75-75correct
D−84-84
Solution
Step 1: sin⁡2x=1−cos⁡2x\sin^2 x=1-\cos^2 x, c=cos⁡x∈[−1,1]c=\cos x\in[-1,1]:
3(1−c2)+12c−3=−3c2+12c=−3((c−2)2−4)=12−3(c−2)2=p.3(1-c^2)+12c-3=-3c^2+12c=-3\big((c-2)^2-4\big)=12-3(c-2)^2=p.
 Step 2: c∈[−1,1]⇒(c−2)∈[−3,−1]⇒(c−2)2∈[1,9]c\in[-1,1]\Rightarrow(c-2)\in[-3,-1]\Rightarrow(c-2)^2\in[1,9]:
pmax⁡=12−3(1)=9,pmin⁡=12−3(9)=−15.p_{\max}=12-3(1)=9,\qquad p_{\min}=12-3(9)=-15.
∴ p∈[−15,9].\therefore\ p\in[-15,9].
 Step 3: ∑p=−159p=−(10+11+12+13+14+15)=−75\displaystyle\sum_{p=-15}^{9}p=-(10+11+12+13+14+15)=-75. Correct answer: (3)
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