Question

How do you solve `3sin( 2 x ) +cos( 2 x ) = 0` from 0 to 2pi?

Answer 1

`S={1.4099,2.9807,4.5515,6.1223}`

Explanation:

Use the following formulas to Transform the Equation
`Acosx+Bsinx=C cos (x-D)`
`C=sqrt(A^2+B^2), cosD=A/C, sinD=B/C`

Note that cosine and sine are both positive which means quadrant one

`A=1, B=3, C=sqrt(1^2+3^2)=sqrt10`

`cos D=1/sqrt10->D=cos^-1(1/sqrt10)=1.249...`

`sqrt10 cos(2x-1.249....)=0`-> make sure to carry all the digits until the end then round

`cos(2x-1.249...)=0`

`2x-1.249...=cos^-1(0)`

`2x-1.249... = pi/2 +pin`

`2x=1.249+pi/2 +pin`

`2x=2.8198...+pin`

#x=1.4099...+pi/2 n#

`n=0, x=1.4099`

`n=1, x= 2.9807`

`n=2, x=4.5515`

`n=3, x=6.1223`

`n=4, x=7.6931>2pi` therefore we stop

`S={1.4099,2.9807,4.5515,6.1223}`

Answer 2

`-9^@22 (or 350^@78); 80^@79 ; 170^@79`

Explanation:

Divide both sides by 3.
`sin (2x) + (1/3)cos (2x) = 0`
Call `tan t = (sin t)/(cos t) = 1/3` --> `t = 18^@43.`
The equation becomes:
`sin 2x .cos t + sin t.cos 2x = 0`
Apply the trig identity: sin (a + b) = sin a.cos b + sin b.cos a
`sin (2x + t) = 0`
a. sin 2x + 18.43 = 0 --> 2x = - 18.43 --> `x = -9^@22`
`sin (2x + t) = pi = 180`. --> 2x = 180 - 18.43 = 161.57 --> `x = 80^@79`
`sin 2x + t = 2pi = 360` --> 2x = 360 - 18.43 = 341.57 -->
`x = 170^@79`
Answers for (0, 360)
Note that the co-terminal arc of (-9.22) is arc `(350^@78)`
`80^@79; 170^@79; 350^@78`
Check by calculator.
x = 80.79 --> 2x = 161.68 --> sin 2x = 0.32 --> 3sin 2x = 0.95.
cos 161/68 = - 0.95. Therefor: 0.95 - 0.95 = 0. OK