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Question: How do you find the general solutions for \(\sin x=-\dfrac{1}{2}\) ? (a) Using trigonometric angle...

How do you find the general solutions for sin⁡x=−12\sin x=-\dfrac{1}{2} ?
(a) Using trigonometric angle identities
(b) Using linear formulas
(c) a and b both
(d) None of these

Explanation

Solution

In this problem we are to find the general solutions for sin⁡x=−12\sin x=-\dfrac{1}{2}. We will try to use the trigonometric angle identities to find and simplify the value of our needed problem. We can start with the fact that sin⁡(−θ)=−sin⁡(θ)\sin \left( -\theta \right)=-\sin \left( \theta \right) and consider sin⁡30∘=12\sin 30{}^\circ =\dfrac{1}{2} to get ahead with the problem and evaluate the value. Also to find the general value we will use the property, from, sin a = sin b, we are getting, a=nπ+(−1)n(b)a=n\pi +{{(-1)}^{n}}\left( b \right),

Complete step by step solution:
According to the question, we are to find the general solutions for sin⁡x=−12\sin x=-\dfrac{1}{2}.
Now, as per the trigonometric identities, sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x ,
Again, the trigonometric table of special arcs gives , sin⁡30∘=12\sin 30{}^\circ =\dfrac{1}{2}
From which, we can say, sin⁡(−30∘)=−sin⁡30∘=−12\sin \left( -30{}^\circ \right)=-\sin 30{}^\circ =-\dfrac{1}{2}
So, we get, sin⁡θ=sin⁡(−30∘)=sin⁡(−π6)\sin \theta =\sin \left( -30{}^\circ \right)=\sin \left( -\dfrac{\pi }{6} \right)
Now, the value of a sine function is positive in the first and second quadrant. As the value is negative here, we get, the value would be in the third or fourth quadrant.
Thus, we also getting, 2π−π6=11π62\pi -\dfrac{\pi }{6}=\dfrac{11\pi }{6},
Thus, from the general solution of sin a = sin b, we are getting, a=nπ+(−1)n(b)a=n\pi +{{(-1)}^{n}}\left( b \right),
Hence, we are getting,θ=nπ+(−1)n(−π6)\theta =n\pi +{{(-1)}^{n}}\left( -\dfrac{\pi }{6} \right) where n is an integer.
So, the solution is, (a) Using trigonometric angle identities.

Note: To understand how the values of trigonometric ratios like sin⁡(−π6)\sin \left( -\dfrac{\pi }{6} \right) change in different quadrants, first we have to understand ASTC rule. The ASTC rule is nothing but the "all sin tan cos" rule in trigonometry. The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant. In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.