Answer:
[tex] 56.4^\circ [/tex] and [tex] 123.6^\circ [/tex]
Step-by-step explanation:
To solve the equation [tex]6\sin(\theta) - 5 = 0[/tex] for values of [tex] \theta [/tex] where [tex]0^\circ \leq \theta < 360^\circ[/tex], we need to isolate [tex] \sin(\theta) [/tex] and then find the angles within the given range where this equation holds true.
Starting with the given equation:
[tex] 6\sin(\theta) - 5 = 0 [/tex]
Add 5 to both sides:
[tex] 6\sin(\theta) = 5 [/tex]
Now, divide both sides by 6:
[tex] \sin(\theta) = \dfrac{5}{6} [/tex]
To find [tex] \theta [/tex], we'll use the inverse sine function (arcsin). Remember that the arcsine function returns angles between [tex] -90^\circ [/tex] and [tex] 90^\circ [/tex]. However, we need to be careful to select the angles within [tex]0^\circ \leq \theta < 360^\circ[/tex].
Using a calculator, find the arcsin of [tex] \dfrac{5}{6} [/tex]:
[tex] \arcsin\left(\dfrac{5}{6}\right) \approx 56.44269024 \approx 56.4^\circ [/tex]
Since [tex] \sin(\theta) [/tex] is positive in both the first and second quadrants, we have another solution in the second quadrant.
To find this, we'll subtract [tex] 56.4^\circ [/tex] from [tex] 180^\circ [/tex] to find the angle in the second quadrant:
[tex] 180^\circ - 56.4^\circ \approx 123.6^\circ [/tex]
So, the solutions for [tex] \theta [/tex] within the range [tex] 0^\circ \leq \theta < 360^\circ [/tex] are approximately [tex] 56.4^\circ [/tex] and [tex] 123.6^\circ [/tex].
Answer:
θ = 56.4°
θ = 123.6°
Step-by-step explanation:
Given trigonometric equation:
[tex]6 \sin \theta - 5 = 0[/tex]
To solve the equation for all values of θ, such that 0° ≤ θ < 360°, begin by isolating sin θ:
[tex]6 \sin \theta - 5 +5= 0+5\\\\\\6 \sin \theta=5\\\\\\\dfrac{6 \sin \theta}{6}=\dfrac{5}{6}\\\\\\\sin \theta=\dfrac{5}{6}[/tex]
Now, take the inverse sine:
[tex]\theta=\sin^{-1}\left(\dfrac{5}{6}\right)\\\\\\\theta=56.442690238...^{\circ}\\\\\\\theta=56.4^{\circ}\; \sf (nearest\;tenth)[/tex]
Sine is positive in quadrants I and II of the unit circle. As we need to ensure that we find solutions within the specified range of 0° ≤ θ < 360°, and the first angle we have found is in quadrant I, we also need to consider the angle in quadrant II that has the same sine value. This angle can be found by subtracting the result from 180°:
[tex]\theta=180^{\circ} - 56.442690238...^{\circ}\\\\\\\theta=123.55730976...^{\circ}\\\\\\\theta=123.6^{\circ}\; \sf (nearest\;tenth)[/tex]
So, the solutions for all values of θ, such that 0° ≤ θ < 360°, are:
[tex]\Large\boxed{\boxed{56.4^{\circ}}}[/tex]
[tex]\Large\boxed{\boxed{123.6^{\circ}}}[/tex]
