Question

2 sin(5x+4)=-square root 3

Answer 1

Do you know how to solve 2sin(u)-sqrt(3) for u?

Answer 2

there is suppose to be an equal sign there somewhere

Answer 3

\[2 \sin(u)=-\sqrt{3}\]

Answer 4

yes

Answer 5

ok then solve that equation first then replace u with 5x+4
then solve for x

Answer 6

why should you cancel out first the 5x+4?

Answer 7

no one is canceling it

Answer 8

uhm. i mean why should we forst get the value of 2sin=-square root 3?

Answer 9

y=sin(x) and y=sin(5x+4) is basically the same graph except y=sin(x) is y=sin(5x+4)'s parent. y=sin(5x+4) is a transformation of the graph y=sin(x).

Answer 10

oh. but should we give two cases of x since there are two values of u?

Answer 11

Example:
2sin(x+1)=1
sin(x+1)=1/2
We don't even really have to replace x+1 with u. I just thought it would be easier for you.
sin is 1/2 when x+1=pi/6+2npi or when x+1=5pi/6+2npi
so we have x=pi/6+2npi-1 or when x=5pi/6+2npi-1

Answer 12

y=sin(x+1) is just the graph of y=sin(x) moved to the left 1 unit

Answer 13

That is why we got the answer answers as we would have gotten for sin(x) except -1

Answer 14

why is there a 2npi?

Answer 15

n*2pi
2pi is the circumference around a circle with radius 1
n represents how many circle rotations we make

Answer 16

do we need to get the value of sin =1/2 first?

Answer 17

sin(u)=1/2? if we are looking at my example yes

Answer 18

last question.2cos^2(3x)+5cos(3x)-3=0 how about this?

Answer 19

That is a quadratic in terms of cos(3x)

Answer 20

just do the same? but many cases?

Answer 21

solve the quadratic cos(3x) first
\[2u^2+5u-3=0\]

Answer 22

if it helps you can replace the cos(3x) with u first
if it looks less nasty to you
i did that above just now

Answer 23

yeah I got it now thank you very much

Answer 24

but if you use the general addition formula.... would the answer be the same?

Answer 25

\[2\sin^2(x)+5\sin(x)-3=0 \\2\sin^2(x)+6\sin(x)-1\sin(x)-3=0 \\ 2\sin(x)[\sin(x)+3]-1[\sin(x)+3]=0 \\ (\sin(x)+3)(2\sin(x)-1)=0 \\ \text{ we have two equations \to solve } \sin(x)+3=0 \text{ or } 2\sin(x)-1=0 \\sin(x)=-3 \text{ or } \sin(x)=\frac{1}{2} \\ \text{ first off } \sin(x) \neq -3 \text{ since }\sin(x) \in [-1,1] \\ \sin(x)=\frac{1}{2} \text{ when } x=\frac{\pi}{6}+2npi \text{ or } x=\frac{5\pi}{6}+2npi\]

Answer 26

That is just another example above having a quadratic in terms of a trig function

Answer 27

what general addition formula?

Answer 28

oh nevermind lol. just got it

Answer 29

thanks!

Answer 30

np