Equations and Graphs of Trigonometric Functions. Determine the solutions to the equation 4sin2(x+45)=3 over the interval 0<x<360 Algebreically?

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Answer 1 · 2018-07-27T13:17:36

$x \approx {20.7}^{\circ} and x \approx {339.3}^{\circ}$ $x ~~ 20.7^@ and x ~~ 339.3^@$

We have the equation

$4 sin 2 (x + 45^{\circ}) = 3$ $4sin 2(x+45^@) = 3$
$4 sin (2 x + 90^{\circ}) = 3$ $4 sin(2x+90^@) = 3$

Remember the sum identity for sine:

$sin (a + b) = sin a cos b + cos a sin b$ $sin(a+b) = sinacosb+cosasinb$

$\Rightarrow sin (2 x + 90^{\circ}) = sin 2 x {cos 90}^{\circ} + cos 2 x {sin 90}^{\circ} =$ $=> sin(2x+90^@) = sin2xcos90^@+cos2xsin90^@=$

$= sin 2 x \cdot 0 + cos 2 x \cdot 1 = cos 2 x$ $=sin2x * 0 + cos2x * 1 = cos2x$

$:. 4 cos2x = 3=> cos2x = 3/4$

Now, there is no nice form of the answer. We might as well accept that

$2x = arccos(3/4)$

However, as the inverse cosine goes from $0^@$ to $180^@$ , these are not all the solutions; A more general solution would be

$2x=arccos(3/4)+360^@n$ , for $n in ZZ$ .

Even then, since the cosine function is even, meaning that

$cos(-2x) = cos(2x) = 3/4$

We reach the additional solutions:

$2x = +- arccos(3/4) + 360^@n$

Out of the complete set of solutions, only two are located in the interval $(0^@, 360^@)$ :

$x = arccos(3/4) and x = -arccos(3/4) + 360^@$

$arccos(3/4) ~~ 20.7^@ => -arccos(3/4) + 360^@ = 339.3^@$