Question

How do you solve $2{{\cos }^{2}}x-\sin x-1=0$?

Answer 1

Hint: In this question, we are given an equation in the form of trigonometric function. We need to find the values of x. For this, we will first use ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to change ${{\cos }^{2}}x$ in terms of ${{\sin }^{2}}x$. After that we will equate sinx as a so that we get a quadratic equation which is easy to solve. Then we will solve the equation using split the middle term method and find values of a = sinx. Then we will find the value of x for which the value of sinx equals using trigonometric ratio tables. At last, we will use that value of x to find the general value of x. We will use $\sin x+\sin y\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}y$ where $n\in z$.

Complete step by step answer:
Here we are given an equation as $2{{\cos }^{2}}x-\sin x-1=0$. We need to find the value of x.
Let us first change the ${{\cos }^{2}}x$ in terms of ${{\sin }^{2}}x$.
We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. So rearranging it we can say ${{\cos }^{2}}x=1-{{\sin }^{2}}x$. Putting this in given equation we get $2\left( 1-{{\sin }^{2}}x \right)-\sin x-1=0\Rightarrow 2-2{{\sin }^{2}}x-\sin x-1=0$.
Rearranging and simplifying the equation we get $2{{\sin }^{2}}x+\sin x-1=0$.
We get a quadratic equation in terms of sinx. To solve it easily, let us suppose that, ${{\sin }^{2}}x=a$. So our equation becomes $2{{a}^{2}}+a-1=0$.
Let us solve it using the split middle term method. We need to split 'a' into two numbers ${{n}_{1}}\text{ and }{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=1\text{ and }{{n}_{1}}\cdot {{n}_{2}}=\left( 2 \right)\left( -1 \right)=-2$.
We know 2-1 = 1 and 2(-1) = -2, so ${{n}_{1}}=2\text{ and }{{n}_{2}}=-1$.
Splitting the middle term we get $2{{a}^{2}}+2a-a-1=0$.
Taking 2a common from the first two terms and -1 from last two terms we get $2a\left( a+1 \right)-1\left( a+1 \right)=0$.
Taking (a+1) common we get $\left( a+1 \right)\left( 2a-1 \right)=0$.
Therefore a+1 = 0 and 2a-1 = 0, $a=-1\text{ and }a=\dfrac{1}{2}$.
Substituting values of a as sinx we get $\sin x=-1\text{ and }\sin x=\dfrac{1}{2}$.
(1) sinx = -1
We know $\sin \dfrac{\pi }{2}=1$. But we need negative of 1, sin function is negative in third quadrant so $\sin \left( \pi +\dfrac{\pi }{2} \right)=-\sin \dfrac{\pi }{2}=-1\Rightarrow \sin \dfrac{3\pi }{2}=-1$.
Therefore $\sin x=\sin \dfrac{3\pi }{2}$.
We know when $\sin x+\sin y\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}y$,$n\in z$. So $x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{3\pi }{2}$,$n\in z$.
(2) $\sin x=\dfrac{1}{2}$.
We know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$ from trigonometric ratio table so $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$.
As $\sin x+\sin y\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}y$,$n\in z$.
So $\sin x=\sin \dfrac{\pi }{6}\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$,$n\in z$.
Therefore the value of x are $n\pi +{{\left( -1 \right)}^{n}}\dfrac{3\pi }{2}\text{ and }n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$,$n\in z$.

Note:
Students should always find the general value of x when interval is not given. While splitting the middle term take care of signs. Note that, putting the value of n will give us different values of x which will satisfy the given equation.