Find a possible solution to the equation sin (3x+13)=cos(4x)

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Find a possible solution to the equation sin (3x+13)=cos(4x)

Geometry
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its part of my geometry course @LegendarySadist trig is one of the sections lol

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nah at my schools it goes algebra 1 geometry algebra 3 and pre calc or discrete math im in geometry summer school
*algebra 2
use a co-function identity
for example, recall one of the co-function identities is: \[\cos(u)=\sin(\frac{\pi}{2}-u)\]
wow that really confused me
why does it confuse you?
because this stuff is like another language to me
but why does the identity confuse you?
is it because you never seen it?
kinda this is my first time learning this
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could you find cos(theta) and sin(90-theta) using this right triangle ?
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\[\sin(90-\theta)=\frac{v}{c} \text{ and } \cos(\theta)=\frac{v}{c} \\ \text{ that means } \sin(90-\theta)=\cos(\theta) \\ \] of course this is all true if we are working in degress
since I used 90 deg there
but we can do the same thing for radians and say \[\sin(\frac{\pi}{2}-\theta)=\cos(\theta)\]
would the answer be 13?
\[\sin(3x+13)=\cos(4x) \\ \text{ by cofunction identity you have that you can find a solution from } \\ \text{ setting the insides of this thingy equal } \\ \sin(3x+13)=\sin(\frac{\pi}{2}-4x)\]
so what would that make the answer be?
set the insides equal
and solve for x
\[3x+13=\frac{\pi}{2}-4x \]
I still don't get my answer
to solve a linear equation first you do: you put all your x terms on one side and your non-x terms on the opposing side by adding and subtracting stuff on both sides
oh so its 11
your answer should have pi in it somewhere so 11 isn't the answer
or a answer in this case
In general to solve the linear equations of the form ax+b=cx+d you do: \[ax+b=cx+d \\ \text{ subtract} cx \text{ on both sides } ax-cx+b=cx-cx+d \\ \text{ now } ax-cx=(a-c)x \text{ and } cx-cx=0 \\ \text{ so we have } (a-c)x+b=0+d \\ \text{ now subtract } b \text{ on both sides } (a-c)x+b-b=d-b \\ (a-c)x+0=d-b \\ (a-c)x=d-b \\ \text{ now final step is to divide }(a-c) \text{ on both sides } \\ \text{ so we have } x=\frac{d-b}{a-c}\]
the options are 11 -13 13 and 0
oh then you didn't have to find a possible solution using any identities just plug in those numbers to see which gives you the same thing on both sides
so what would the answer come out to be
have you tried pluggin in the choices as I suggested above?
i only have 1 minute until i have to submit it
I think they mean all of that to be in degrees so put your calculator on degrees and see which of the following is true: \[\sin(3(0^o)+13^o)=\cos(4(0^o) ) \\ \sin(3(11^o)+13^o)=\cos(4(11^o)) \\ \sin(3(13^o)+13^o)=\cos(4(13^o)) \\ \sin(3(-13^o)+13^o)=\cos(4(-13^o))\]
you can definitely rule out the first since we know sin(13 deg) isn't cos(0 deg) then check the second equation with 11 degs and so on... until you have the same thing on both sides
what is sin(3*11 deg+13 deg)=?
give you a hint 3*11+13=33+13=46 and guess what 90-46 is?
recall the above you can do this part without calculator come on you know sin(x)=cos(90-x) so sin(46)=cos( ? )
also it would be helpful in the future if you say if we are working in deg or radians because I thought it was radians until I seen your choices
good luck

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