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Question: How do you solve \( 2{\sin ^2}x + \sin x = 1 \) for x in the interval \( [0,2\pi ) \) ?...

How do you solve 2sin⁡2x+sin⁡x=12{\sin ^2}x + \sin x = 1 for x in the interval [0,2π)[0,2\pi ) ?

Explanation

Solution

Hint : In the question, we are actually given a polynomial equation of one variable, it is a function of sin⁡x\sin x .When we replace sin⁡x\sin x in the given equation with any other variable, it becomes easier to solve the polynomial equation. We know that the highest power of the variable used in the polynomial is the degree of a polynomial equation and also the number of roots of a polynomial equation is equal to its degree, so the given equation will have two roots that can be found by several methods like factoring the equation or by a special formula called quadratic formula or by completing the square method; if we are not able to factorize the equation we use other methods. So, we can find the values of x after finding the possible values of sin⁡x\sin x .

Complete step-by-step answer :
We have to solve the equation 2sin⁡2x+sin⁡x=12{\sin ^2}x + \sin x = 1
Let sin⁡x=t\sin x = t
⇒2t2+t=1 ⇒2t2+t−1=0   \Rightarrow 2{t^2} + t = 1 \\\ \Rightarrow 2{t^2} + t - 1 = 0 \;
We can solve the above equation by factorization to find the value of t, as follows –
2t2+2t−t−1=0 2t(t+1)−1(t+1)=0 ⇒(2t−1)(t+1)=0 ⇒2t−1=0, t+1=0 ⇒t=12, t=−1   2{t^2} + 2t - t - 1 = 0 \\\ 2t(t + 1) - 1(t + 1) = 0 \\\ \Rightarrow (2t - 1)(t + 1) = 0 \\\ \Rightarrow 2t - 1 = 0,\,t + 1 = 0 \\\ \Rightarrow t = \dfrac{1}{2},\,t = - 1 \;
Now replacing t with the original value, we get –
sin⁡x=12, sin⁡x=−1 ⇒sin⁡x=sin⁡π6, sin⁡x=−sin⁡π2    (sin⁡π6=12, sin⁡π2=1)   \sin x = \dfrac{1}{2},\,\sin x = - 1 \\\ \Rightarrow \sin x = \sin \dfrac{\pi }{6},\,\sin x = - \sin \dfrac{\pi }{2}\,\,\,\,(\sin \dfrac{\pi }{6} = \dfrac{1}{2},\,\sin \dfrac{\pi }{2} = 1) \;
Now,
sin⁡π6=sin⁡(π−π6) ⇒sin⁡π6=sin⁡5π6   \sin \dfrac{\pi }{6} = \sin (\pi - \dfrac{\pi }{6}) \\\ \Rightarrow \sin \dfrac{\pi }{6} = \sin \dfrac{{5\pi }}{6} \;
And
\-sin⁡π2=sin⁡(π+π2), −sin⁡π2=sin⁡(2π−π2) ⇒−sin⁡π2=sin⁡3π2   \- \sin \dfrac{\pi }{2} = \sin (\pi + \dfrac{\pi }{2}),\, - \sin \dfrac{\pi }{2} = \sin (2\pi - \dfrac{\pi }{2}) \\\ \Rightarrow - \sin \dfrac{\pi }{2} = \sin \dfrac{{3\pi }}{2} \;
Thus,
sin⁡x=sin⁡π6=sin⁡5π6=sin⁡3π2 ⇒x=π6=5π6=3π2   \sin x = \sin \dfrac{\pi }{6} = \sin \dfrac{{5\pi }}{6} = \sin \dfrac{{3\pi }}{2} \\\ \Rightarrow x = \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} = \dfrac{{3\pi }}{2} \;
Hence, the solution of the equation 2sin⁡2x+sin⁡x=12{\sin ^2}x + \sin x = 1 is x=π6x = \dfrac{\pi }{6} or 5π6\dfrac{{5\pi }}{6} or 3π2\dfrac{{3\pi }}{2} .
So, the correct answer is “ x=π6x = \dfrac{\pi }{6} or 5π6\dfrac{{5\pi }}{6} or 3π2\dfrac{{3\pi }}{2} ”.

Note : The value of the sine function is positive in the first and the second quadrant that’s why we have got 3 answers to the above equation. We can find infinite answers as the value of x but we are given that the solution lies in the interval [0,2π)[0,2\pi ) that is the value of x can be greater than or equal to zero but smaller than 2π2\pi so we take only these three values as the answer.