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Question: How do you solve \( 2{\sin ^2}x + 3\sin x + 1 = 0 \) and find all the solutions in the interval \( [...

How do you solve 2sin⁡2x+3sin⁡x+1=02{\sin ^2}x + 3\sin x + 1 = 0 and find all the solutions in the interval [0,2π)[0,2\pi ) ?

Explanation

Solution

Hint : On replacing sin⁡x\sin x in the given equation with any other variable, we see that the equation is a polynomial equation. Now, the degree of a polynomial equation is the highest power of the variable used in the polynomial. A polynomial equation has exactly as many roots as its degree, so the given equation will have two roots that can be found by factoring the equation or by a special formula called completing the square method if we are not able to factorize the equation. So, after finding the possible values of sin⁡x\sin x , we can find the values of x.

Complete step-by-step answer :
We have to solve the equation 2sin⁡2x+3sin⁡x+1=02{\sin ^2}x + 3\sin x + 1 = 0
Let sin⁡x=t\sin x = t
⇒2t2+3t+1=0\Rightarrow 2{t^2} + 3t + 1 = 0
The equation written above is a quadratic polynomial equation, so 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⁡(2π−π6) ⇒−sin⁡π6=sin⁡7π6, −sin⁡π6=sin⁡11π6   \- \sin \dfrac{\pi }{6} = \sin (\pi + \dfrac{\pi }{6}),\, - \sin \dfrac{\pi }{6} = \sin (2\pi - \dfrac{\pi }{6}) \\\ \Rightarrow - \sin \dfrac{\pi }{6} = \sin \dfrac{{7\pi }}{6},\, - \sin \dfrac{\pi }{6} = \sin \dfrac{{11\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⁡7π6=sin⁡11π6=sin⁡3π2 ⇒x=7π6=11π6=3π2   \sin x = \sin \dfrac{{7\pi }}{6} = \sin \dfrac{{11\pi }}{6} = \sin \dfrac{{3\pi }}{2} \\\ \Rightarrow x = \dfrac{{7\pi }}{6} = \dfrac{{11\pi }}{6} = \dfrac{{3\pi }}{2} \;
Hence, the solution of the equation 2sin⁡2x+3sin⁡x+1=02{\sin ^2}x + 3\sin x + 1 = 0 is x=7π6x = \dfrac{{7\pi }}{6} or 11π6\dfrac{{11\pi }}{6} or 3π2\dfrac{{3\pi }}{2} .
So, the correct answer is “ 7π6\dfrac{{7\pi }}{6} or 11π6\dfrac{{11\pi }}{6} or 3π2\dfrac{{3\pi }}{2} ”.

Note : The standard form of a quadratic polynomial equation is ax2+bx+c=0a{x^2} + bx + c = 0 .While solving an equation by factorization, the condition to form factors is that we have to express the coefficient of x as a sum of two numbers such that their product is equal to the product of the coefficient of x2{x^2} and the constant cc that is a×c=b1×b2a \times c = {b_1} \times {b_2} . 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 .