How do you solve $3 sin x + 5 cos x = 4$ $3sinx+5cosx=4$ ?

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How do you solve $3 sin x + 5 cos x = 4$ $3sinx+5cosx=4$ ?

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Answer 1 · 2016-11-30T15:17:03

Pose:

$sin x = u$ $sinx = u$
$cos x = v$ $cos x = v$

You have:

$3 u + 5 v = 4$ $3u+5v=4$

and from the fundamental trigonometric identity:

$u^{2} + v^{2} = 1$ $u^2+v^2=1$

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

$v = \frac{4 - 3 u}{5}$ $v=(4-3u)/5$

$u^{2} + {(\frac{4 - 3 u}{5})}^{2} = 1$ $u^2+((4-3u)/5)^2=1$

$25 u^{2} + 16 - 24 u + 9 u^{2} - 25 = 0$ $25u^2+16-24u+9u^2-25=0$

$34 u^{2} - 24 u - 9 = 0$ $34u^2-24u-9=0$

Using the quadratic formula:

$u = \frac{12 \pm \sqrt{144 + 9 \cdot 34}}{34} = \frac{12 \pm 3 \sqrt{50}}{34}$ $u=frac (12+-sqrt(144+9*34)) 34=frac (12+-3sqrt(50)) 34$

$u_{1} = 0.97685$ $u_1=0.97685$
$u_{2} = - 0.27097$ $u_2=-0.27097$

$v_{1} = 0.21389$ $v_1=0.21389$
$v_{2} = 0.96258$ $v_2=0.96258$

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

The corresponding angles are:

$θ_{1} = - 0.274407416$ $theta_1=-0.274407416$
$θ_{2} = 1.355246417$ $theta_2=1.355246417$

Answer 2 · 2016-11-30T17:00:14

$x = arcsin (\frac{4}{\sqrt{3^{2} + 5^{2}}}) - arctan (\frac{5}{3})$ $x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)$

Explanation:

Making

$3 = λ cos (ϕ_{0})$ $3=lambda cos(phi_0)$

and

$5 = λ sin (ϕ_{0})$ $5=lambda sin(phi_0)$

with $λ = \sqrt{3^{2} + 5^{2}}$ $lambda = sqrt(3^2+5^2)$

we have

$λ sin (x) cos (ϕ_{0}) + λ cos (x) sin (ϕ_{0}) = 4$ $lambda sin(x) cos(phi_0)+lambda cos(x)sin(phi_0)=4$

Now using the identity

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

$sin (x + ϕ_{0}) = \frac{4}{λ} = \frac{4}{\sqrt{3^{2} + 5^{2}}}$ $sin(x+phi_0)=4/lambda=4/sqrt(3^2+5^2)$

so $x + ϕ_{0} = arcsin (\frac{4}{\sqrt{3^{2} + 5^{2}}})$ $x+phi_0=arcsin(4/sqrt(3^2+5^2))$

but $ϕ_{0} = arctan (\frac{5}{3})$ $phi_0 = arctan(5/3)$ then

$x = arcsin (\frac{4}{\sqrt{3^{2} + 5^{2}}}) - arctan (\frac{5}{3})$ $x=arcsin(4/sqrt(3^2+5^2))-arctan(5/3)$

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