How do you solve #3sinx+5cosx=4#?

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Answer 1 · 2016-11-30T15:17:03

Pose:

#sinx = u#
#cos x = v#

You have:

#3u+5v=4#

and from the fundamental trigonometric identity:

#u^2+v^2=1#

Substitute #v# from the first equation into the second:

#v=(4-3u)/5#

#u^2+((4-3u)/5)^2=1#

#25u^2+16-24u+9u^2-25=0#

#34u^2-24u-9=0#

Using the quadratic formula:

#u=frac (12+-sqrt(144+9*34)) 34=frac (12+-3sqrt(50)) 34#

#u_1=0.97685#
#u_2=-0.27097#

#v_1=0.21389#
#v_2=0.96258#

Both solutions are between #-1# and #1# as the value of a sine or a cosine must be and can be accepted.

The corresponding angles are:

#theta_1=-0.274407416#
#theta_2=1.355246417#

Answer 2 · 2016-11-30T17:00:14

#x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)#

Making

#3=lambda cos(phi_0)#

and

#5=lambda sin(phi_0)#

with #lambda = sqrt(3^2+5^2)#

we have

#lambda sin(x) cos(phi_0)+lambda cos(x)sin(phi_0)=4#

Now using the identity

#sin(a+b)=sin(a)cos(b)+sin(b)cos(a)# we have

#sin(x+phi_0)=4/lambda=4/sqrt(3^2+5^2)#

so #x+phi_0=arcsin(4/sqrt(3^2+5^2))#

but #phi_0 = arctan(5/3)# then

#x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)#