Question

sinx + sin2x + sin3x = cosx + cos2x + cos 3x. find x

Answer 1

do u know what CosA+cosB =?
and what SinA+sinB =?

Answer 2

i know, but my teacher said that i had to use problem of complex numbers,

Answer 3

iam getting 5 values for x

Answer 4

one of them is coming out to be \[\huge \frac{ 2 }{ 3 }(3\pi n-\pi)\]

Answer 5

what i got is
\[\cos(x) = -\frac{ 1 }{ 2 }\]and
\[\sin{2x} = \cos{2x}\]
Just two equatiions @deena123 what did u get?

Answer 6

iam getting 5 solutions for x !

Answer 7

I just got two as you can see ?. what are other three? Can u write your procedure please?

Answer 8

is the general formula for this question--->\[\huge \frac{ \pi }{ 8 }+n \frac{ \pi }{ 2 }\]

Answer 9

2 sin2x .cosx +sin 2x = 2 cos2x. cosx + cos2x

sin2x(2 cosx+ 1) = cos2x( 2 cosx + 1)

sin2x = cos 2x

sin2x/cos2x =1

tan2x = 1

2x = tan^-1 (1)

2x =pi/4

x = pi/8. ( a particular solution)

hence general solution : x = pi/8 +n pi/2.