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solve cot x + 6 sin x - 2 cos x =3

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solve cot x + 6 sin x - 2 cos x =3

Jun 27, 2017

#1
+98130
+5

cot x + 6 sin x - 2 cos x =3     we can write

cosx / sinx  +  6sin x - 2 cosx  = 3       multiply through by sin x

cos x  +  6 sin^2x  -  2cosx sin x  =  3sin x

cos x + 6sin^2x  - 2sin cos x  - 3sin x  = 0         we can factor this as

( 3sinx  - cosx) (2sinx  - 1) = 0      set each factor to 0  and solve

3sinx  - cos x  = 0

3sin x  = cos x

3sinx / cos x  =  1

3 tan x  =  1

tan x  = 1/3

arctan (1/3)  ≈  18.435°  + 180°n    where  n is an intreger

For the other factor we have that

2sinx - 1  = 0

2sin x  = 1

sin x  =  1/2

arcsin (1/2)    and this happens at  30° + 360°n    and at  150° + 360°n

And again, n is an integer

Here's ther graph : https://www.desmos.com/calculator/eiai1cs0i1

Jun 27, 2017
#2
+7352
+4

Nice CPhill  !! This was harder than it first looks, for me at least !  Here is just the factoring part. I did it at first just to see it for myself, and then I figured it couldn't hurt to post it.

cos x  +  6 sin2x  -  2 sin x cos x  -  3 sin x   =   0        Rearrange.

6 sin2x  -  3 sin x  +  cos x  -  2 sin x cos x   =  0         Factor out  3 sin x  from the first two terms and

factor out  -cos x  from the last two terms.

3 sin x (2 sin x - 1)  -  cos x (-1 + 2 sin x)  =  0             Factor out  (2 sin x - 1)  from both terms.

(2 sin x - 1)(3 sin x - cos x)  =  0

Jun 27, 2017