Question

Solve 2cos(x)-4sin(x)=3 [0,360]

Answer 1

cos(x)^2 + sin(x)^2 = 1
so
cos(x) = [1 -sin(x)^2]^0.5

plug it into the equation:
2[1-sin(x)^2]^0.5 - 4sin(x) = 3
2[1-sin(x)^2]^0.5 = 3 + 4sin(x)

square both sides

4[1-sin(x)^2] = 9 + 24sin(x) + 16sin(x)^2

can you do it now ?

Answer 2

can you do it using harmonic form

Answer 3

another way to do this is by complementary angle method
angle A

R cos (x + A) = R cosx cosA - R sinx x sin A = 1

Answer 4

mm
Rcos(x+a) = Rcos(x)cos(a) - Rsin(x)sin(a)

Rcos(a) = 2 , Rsin(a) = 4
-> tan(a) = 2 -> a = 63.43
so R = 4.47

Answer 5

I found \[2\sqrt{5}\cos(\theta+116.6)-3=0\]

Answer 6

\[\text{this is how i find} \\ \\ 2\cos \theta-4\sin \theta=3~~~~~~~~~Rcos(\theta+\alpha) \\ \\ 2\cos \theta-4\sin \theta -3=0 ~~~~~~~Rcos \theta \cos \alpha - R \sin \theta \sin \alpha \\ \\ \cos \theta (2-Rcos \alpha)-\sin \theta (4+\sin \alpha ) -3=0 \\ \\ Rcos \alpha=2 ~~~~~~~Rsin \alpha=-4 \\ \\ \tan \alpha =-2 \\ \\ \alpha=116.6 \\ \\R=\sqrt{4+16} \\ \\R=2\sqrt{5} \\ \\ 2\sqrt{5}\cos( \theta +116.6)-3=0\]

Answer 7

oh crap i made a mistake...
i carried on oops

2cos(x)-4sin(x)=3
2cos(x)-4sin(x) is equivalent to Rcos(x+a)
tana=4/2=2
a=tan^-1(2)
R=root(a²+b²)
R=root 20
R=2root5
2root5Cos(x+63.4)=3

now divide by 2root5 on both sides

cos(x+63.4)=3/2root5

inverse cos

x+63.4=cos^-1(3/2root5)

that should help you

Answer 8

are you sure the angle is 63.4

Answer 9

Tan(a)=-b/a
b=-4
a=2
--4/2=4/2=2
tanA=2
inverse tan(2)=63.4

Answer 10

does that help you?