Answer:
The solution set is { 66.6°, 113.4°, 192.6°, 347.4° }.
Step-by-step explanation:
Given:
The given equation in the interval (0°,360°) is given as:
[tex]10\sin ^{2}\theta - 7\sin \theta=2[/tex]
Adding -2 both sides, we get
[tex]10\sin ^{2}\theta - 7\sin \theta-2=2-2\\10\sin ^{2}\theta - 7\sin \theta-2=0[/tex]
This is a quadratic equation in [tex]\sin \theta[/tex] of the form [tex]ax^2+bx+c=0[/tex]. Solve it using the quadratic formula given as:
[tex]\sin \theta=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
Here, [tex]a=10,b=-7,c=-2[/tex]
Therefore,
[tex]\sin \theta=\frac{-(-7)\pm \sqrt{(-7)^{2}-4(10)(-2)}}{2(-2)}\\\sin \theta=0.918\textrm{ or }\sin \theta = -0.218[/tex]
Now, taking inverse of the sine values, we get
[tex]\theta=\sin^{-1}(0.918)\textrm{ or }\theta=\sin^{-1}(-0.218)\\\theta=66.6\textrm{ or }\theta=-12.59[/tex]
But, we need to find [tex]\theta[/tex] in the interval (0°,360°).
[tex]\textrm{For interval (0,180)}\\\sin\theta=sin(180-\theta)\\\\\textrm{For interval (180,270)}\\\sin\theta=sin(180+\theta)\\\\\textrm{For interval (180,360)}\\\sin\theta=sin(360-\theta)[/tex]
[tex]\theta=-12.59[/tex] is in the fourth quadrant, So, its positive value is:
[tex]360-12.59=347.41[/tex]°
Also, from unit circle, sine of the angle in third and fourth quadrants are same. So,
[tex]12.59=180+12.59=192.59[/tex]°
[tex]66.6=180-66.6=113.4[/tex]°
Therefore, the set of angles for which the above equation is satisfied in the interval (0°,360°) is { 66.6°, 113.4°, 192.6°, 347.4° }.
