How do you find the solution to $sin θ + 3 = 5 sin θ$ $sintheta+3=5sintheta$ if $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2017-07-14T14:05:43

$θ = {0.848}^{r}, {2.294}^{r}$ $theta = 0.848^r, 2.294^r$

We want to solve:

$sin θ + 3 = 5 sin θ$ $sin theta + 3 = 5sin theta$ where $θ \in [0, 2 π)$ $theta in [0,2pi)$

We can write this as:

$\ \ 4sin theta = 3$
$:. sin theta = 3/4$

If we consider the graph of $y=sinx$ then we can see there will be two solutions:

$sin theta = 3/4 =>$
$" " theta_1 = arcsin(3/4)$ (the fundamental value)
$" " \ \ \ = 0.848^r$ (3dp)

So the two solutions in the specified range are:

$theta = theta_1, pi-theta_1$
$\ \ = 0.848^r, 2.294^r$

How do you find the solution to sintheta+3=5sinthetasinθ+3=5sinθsintheta+3=5sintheta if 0<=theta<2pi0≤θ<2π0<=theta<2pi?