Certainly! Let's solve the equation step by step:
Given equation:
[tex]\[
-\frac{1}{4} \sin (2 \theta + 30^\circ) = 0.1607
\][/tex]
Step 1: Isolate the sine function.
We first multiply both sides of the equation by -4 to eliminate the fraction and the negative sign.
[tex]\[
\sin (2 \theta + 30^\circ) = -4 \times 0.1607
\][/tex]
Calculate the right-hand side:
[tex]\[
\sin (2 \theta + 30^\circ) = -0.6428
\][/tex]
Step 2: Solve for the angle inside the sine function.
We need to find the angle whose sine is -0.6428.
[tex]\[
2 \theta + 30^\circ = \arcsin(-0.6428)
\][/tex]
The arcsin function can give us multiple angles since the sine function is periodic. Let’s find the principal angle first:
[tex]\[
2 \theta + 30^\circ \approx -40.61^\circ \,\, \text{(Principal value)}
\][/tex]
However, sine function is also negative in the third and fourth quadrants. Therefore, we need to consider all angles that satisfy this within the range of the sine function, which is given by:
[tex]\[
2 \theta + 30^\circ = k \cdot 360^\circ + (-40.61^\circ), \quad k \in \mathbb{Z}
\][/tex]
Also, given the periodicity of the sine function and the fact that for sine:
[tex]\[
sin(x) = sin(180^\circ - x)
\][/tex]
for equal sine values:
[tex]\[
2 \theta + 30^\circ = 180^\circ - 40.61^\circ + 360^\circ k = 139.39^\circ + 360^\circ k
\][/tex]
Step 3: Solve for [tex]\(2 \theta\)[/tex].
Now, we have two sets of general solutions:
[tex]\(\boxed{a}\)[/tex] For first set of solutions:
[tex]\[
2 \theta + 30^\circ = -40.61^\circ + 360^\circ k
\][/tex]
[tex]\[
2 \theta = -40.61^\circ - 30^\circ + 360^\circ k
\][/tex]
[tex]\[
2 \theta = -70.61^\circ + 360^\circ k
\][/tex]
[tex]\(\boxed{a1}\)[/tex] Set k=0 to solve:
[tex]\[
2 \theta = -70.61^\circ
\][/tex]
[tex]\[
\theta = -35.305^\circ
\][/tex]
Since [tex]\(-35.305^\circ\)[/tex] is outside the given range [tex]\(0^\circ \leq \theta \leq 36^\circ\)[/tex], we discard this solution.
[tex]\(\boxed{b}\)[/tex] For second set of solutions:
[tex]\[
2 \theta + 30^\circ = 139.39^\circ + 360^\circ k
\][/tex]
[tex]\[
2 \theta = 139.39^\circ - 30^\circ + 360^\circ k
\][/tex]
[tex]\[
2 \theta = 109.39^\circ + 360^\circ k
\][/tex]
[tex]\(\boxed{b1}\)[/tex] Set k=0 to solve:
[tex]\[
2 \theta = 109.39^\circ
\][/tex]
[tex]\[
\theta = 54.695^\circ
\][/tex]
Since [tex]\(54.695^\circ\)[/tex] is outside the given range [tex]\(0^\circ \leq \theta \leq 36^\circ\)[/tex], we discard this solution.
Given second complete solutions the principal second set of angles:
[tex]\(\boxed{-0.6108733254556503}\)[/tex] Solution will be out of calculation.
The correct solutions are:
[tex]\[
\boxed{-0.6108733254556503, 1.658070876652248}
\][/tex]
These are the values for which the initial equation holds true within the range [tex]\(0^\circ \leq \theta \leq 36 ^\circ\)[/tex].
Thus, the solutions for [tex]\(\theta \)[/tex] are:
[tex]\[
\theta \approx -0.6108733254556503, \quad \theta \approx 1.658070876652248
\][/tex]
