0

Guest

Courses New

1.Prove that sin(3x) = 4sinx.sin2x.sin4x 2.Solve : cosx(1+sin x/4 - 2cosx) + sinx (cosx/4 - 2sinx) = 0 3.solve : cos 3x + cos 2x = sin 3x/2 + sin x/2 , x=(0,pi) 4.solve for x and y : cos 2 x + cosx.cosy + cos 2 y = 0 Please solve quickly

1.Prove that sin(3x) = 4sinx.sin2x.sin4x


2.Solve :  cosx(1+sin x/4 - 2cosx) + sinx (cosx/4 - 2sinx) = 0


3.solve :  cos 3x + cos 2x = sin 3x/2 + sin x/2 , x=(0,pi)


4.solve for x and y :  cos2x + cosx.cosy + cos2y = 0


Please solve quickly

Grade:12

2 Answers

Sham Singh Namdhari
34 Points
7 years ago
4sinx*sin2x*sin4x=sin3x LHS= 2(cosx-cos3x)sin4x= sin5x+sin3x-sin7x-sinx=sin3x So, sin5x-sin7x=sinx or, -2cos6x*sinx=sinx so, either sinx=0 or cos6x=-1/2 so, x= n*pi or x=(2n*pi+/- 2*pi/3)/6 where n is any integer
ayisha
15 Points
4 years ago
2(2sinx.sin2x).sin4x=sin3x
2(cosx-cos3x0.sin4x=sin3x
2cosxsin4x-2cos3xsin4x=sin3x
2sin4xcosx-2sin4xcos3x=sin3x
sin5x+sin3x-(sin7x+sinx)=sinx
sin5x-sin7x=sinx
2cos6xsin(-x)=sinx
-2cos6xsinx=sinx
-2cos6xsinx-sinx=0
sinx=0 or cos6x =-1/2
x=npie

Think You Can Provide A Better Answer ?

Other Related Questions on Trigonometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More