Search for: Solution(s) of the equation sinx2+cosx2=2sinx is/are Solution(s) of the equation sinx2+cosx2=2sinx is/are Ax=nπ+(−1)n+1π32,n∈ZBx=nπ+(−1)n+1π22,n∈ZCx=nπ+(−1)n+1π62,n∈ZDx=2nπ+(−1)nπ22,n∈Z Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:sinx2+cosx2=2sinx⇒1+2sinx2cosx2=2sin2x⇒1+sinx=2sin2x⇒(2sinx+1)(sinx−1)=0⇒sinx=−12 or sinx=1⇒x=nπ+(−1)n−π6 or x=nπ+(−1)nπ2,n∈Z⇒x=nπ+(−1)n+1π62 or x=nπ+(−1)nπ22,n∈ZPost navigationPrevious: The domain of the functionf(x) = Cos-1(sec (cos-1X)) + sin-1 (Cosec(sin-1 X)) isNext: if 2cosx+6sinx=max(sinθ+cosθ),then x= Related content JEE Main 2023 Result: Session 1 NEET 2024 JEE Advanced 2023 NEET Rank Assurance Program | NEET Crash Course 2023 JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria