I will give 100 points and mark your answer as brainiest if you answer these question's correctly
1.Find the general solution of 2cos^2x+3Sinx=0
2.Find the middle term in the expansion of (x/3+9y)^16
3.Differentiate secX,sinX,and CosXwith respect to x from first principle
4.show that (cos7x+cos3x-cos5x-cosx)÷(sin7x-sin3x-sin5x+sinx)=cot2x
Please dont copy from the web and dont write short answers thank you
i will ask another question within the next half hour to give out 100 more points to the person who answers this question correctley
Answers
Step-by-step explanation:
1. 2cos^2x+3sinx=0
2(1-sin^2x)+3sinx=0
2-2sin^2x +3sinx=0
2sin^2x -3sinx -2=0
2sin^2x -4sinx+sinx-2=0
2sin x(sin x-2)+1(sinx-2)=0
(sin x -2)(2sin x+1)=0
sin x - 2=0 or 2 sin x+1=0
sin x =2 or sin x = -1/2
4. cos7x + cos3x - cos5x - cosx)/(sin7x - sin3x - sin5x + sinx)
= [cos(5x+2x) + cos(5x-2x) - cos(3x+2x) - cos(3x-2x)]/[sin(5x+2x) - sin(5x-2x) - sin(3x+2x) + sin(3x-2x)]
= (cos5x cos2x - sin5x sin2x + cos5x cos2x + sin5x sin2x - cos3x cos2x + sin3x sin2x - cos3x cos2x - sin3x sin2x)/(sin5x cos2x + cos5x sin2x - sin5x cos2x + cos5x sin2x - sin3x cos2x - cos3x sin2x + sin3x cos2x - cos3x sin2x)
= (2cos5x cos2x - 2cos3x cos2x)/(2cos5x sin2x - 2cos3x sin2x)
= (2cos2x)(cos5x - cos3x) / [(2sin 2x)(cos5x - cos3x)]
= cos2x / sin2x
= cot 2x.
Answer
Please ask one question at a time