![Pi Day. Mathematical constant. March 14th. March Holiday Calendar. Ratio of a circles circumference to its diameter. Constant number Pi. 10524995 Vector Art at Vecteezy Pi Day. Mathematical constant. March 14th. March Holiday Calendar. Ratio of a circles circumference to its diameter. Constant number Pi. 10524995 Vector Art at Vecteezy](https://static.vecteezy.com/system/resources/previews/010/524/995/original/pi-day-mathematical-constant-march-14th-march-holiday-calendar-ratio-of-a-circles-circumference-to-its-diameter-constant-number-pi-vector.jpg)
Pi Day. Mathematical constant. March 14th. March Holiday Calendar. Ratio of a circles circumference to its diameter. Constant number Pi. 10524995 Vector Art at Vecteezy
![prove that, sin2(2pi/7) + sin2(3pi/14) + sin2(11pi/14) + sin2(5pi/7) = 2 - Maths - Trigonometric Functions - 15066307 | Meritnation.com prove that, sin2(2pi/7) + sin2(3pi/14) + sin2(11pi/14) + sin2(5pi/7) = 2 - Maths - Trigonometric Functions - 15066307 | Meritnation.com](https://s3mn.mnimgs.com/img/shared/content_ck_images/ck_43684298bc70078c5977fce9a6bd0f8f.png)
prove that, sin2(2pi/7) + sin2(3pi/14) + sin2(11pi/14) + sin2(5pi/7) = 2 - Maths - Trigonometric Functions - 15066307 | Meritnation.com
Equation sin 7x + cos 2x = 2 Solution is A) x=(2kpi)/7 + 3pi/14,kI B) x=npi + pi/4 , n I C)x= 2npi + pi/2 ,n I D)none of these
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