Question

How do you solve the equation $2{{\sin }^{2}}x-\cos x=1$ when x belongs to the interval $[0,2\pi ]$ .

Answer 1

Now we are given with the equation $2{{\sin }^{2}}x-\cos x=1$ . First we will convert the whole equation in terms of cos with the help of identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . Now once we have converted the whole equation in terms of cos we will substitute cos as t. Now we have a quadratic equation in t. Hence we will solve the equation with the help of formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Now once we have values of t we will substitute the value of t and find x.

Complete step-by-step answer:
Now consider the given equation $2{{\sin }^{2}}x-\cos x=1$
Bringing the terms on RHS to LHS we get,
$2{{\sin }^{2}}x-\cos x-1=0$ .
Now we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .
Hence we can write ${{\sin }^{2}}x=1-{{\cos }^{2}}x$
Substituting this value of ${{\sin }^{2}}x$ in $2{{\sin }^{2}}x-\cos x-1=0$ we get,
$\begin{aligned} & 2\left( 1-{{\cos }^{2}}x \right)-\cos x-1=0 \\\ & \Rightarrow 2-2{{\cos }^{2}}x-\cos x-1=0 \\\ & \Rightarrow -2{{\cos }^{2}}x-\cos x+1=0 \\\ & \Rightarrow 2{{\cos }^{2}}x+\cos x-1=0 \\\ \end{aligned}$
Now let us say cosx = t.
Hence we get,
$2{{t}^{2}}+t-1=0$
Now this is a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ where a = 2, b = 1 and c = -1.
Now the roots of such quadratic equation are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence the roots of this equation $2{{t}^{2}}+t-1=0$ are,
$\begin{aligned} & \Rightarrow \dfrac{-1\pm \sqrt{1-4\left( 2 \right)\left( -1 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow \dfrac{-1\pm \sqrt{1+8}}{4} \\\ & \Rightarrow \dfrac{-1\pm 3}{4} \\\ \end{aligned}$
Hence the roots are $\dfrac{-1-3}{4}=\dfrac{-4}{4}=-1$ and $\dfrac{-1+3}{4}=\dfrac{2}{4}=\dfrac{1}{2}$
Hence we get the possible values of t as – 1 and $\dfrac{1}{2}$
Now since t is nothing but cosx, hence substituting the value of t we get
$\Rightarrow \cos x=-1$ or $\cos x=\dfrac{1}{2}$
Hence $x=\pi$ or $x=\dfrac{\pi }{3}$
Since both $\pi$ and $\dfrac{\pi }{3}$ lie in the interval $[0,2\pi ]$ both are the solution of the given equation.

Note: Now if we consider the equation $2{{t}^{2}}+t-1=0$ we can see that a – b + c = 0 Hence we can say that t = - 1 is one of the roots of the equation. We can find the second root with the help of fact that sum of roots of quadratic $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b}{a}$ . Hence the sum of roots in this case is $\dfrac{-1}{2}$ . Since one root is – 1 we get the second root as $\dfrac{1}{2}$