Question

How do you solve $2{{\sin }^{2}}x-\cos x=1$ on the interval $\left[ 0,2\pi \right]?$

Answer 1

We have to solve $2{{\sin }^{2}}x-\cos x=1$ and we learn what solutions are and then we use ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to simplify our equation then we will change cos x as t to form a quadratic equation and then we will use ways to solve the quadratic function as sin, cos are periodic. So it can have many solutions. So, we only get those that lie in $\left[ 0,2\pi \right].$

Complete step by step answer:
We are given $2{{\sin }^{2}}x-\cos x=1$ and we have to find the solution on $\left[ 0,2\pi \right].$ Solutions are those values which when put in $2{{\sin }^{2}}x-\cos x=1$ will satisfy the equation. To solve our problem, we will first convert the problem into one function. So, as we know
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
So, we get,
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
Using them in $2{{\sin }^{2}}x-\cos x=1$ we get
$\Rightarrow 2\left( 1-{{\cos }^{2}}x \right)-\cos x=1$
$\Rightarrow 2-2{{\cos }^{2}}x-\cos x=1$
Now simplifying, we get,
$\Rightarrow 2{{\cos }^{2}}x+\cos x-1=0$
Now to solve further, let us assume cos x as t, so we will get our equation.
$2{{t}^{2}}+t-1=0$
So we have got our quadratic equation. So we will first simplify it by using a middle term split and then solve it. Now as t can be written as 2t – t, so we get,
$2{{t}^{2}}+t-1=2{{t}^{2}}+\left( 2t-t \right)-1$
On simplifying, we get,
$\Rightarrow 2{{t}^{2}}+t-1=2{{t}^{2}}+2t-t-1$
Taking common in the first two and last two terms, we get,
$\Rightarrow 2{{t}^{2}}+t-1=2t\left( t+1 \right)-1\left( t+1 \right)$
As (t + 1) is the same, so we get,
$\Rightarrow 2{{t}^{2}}+t-1=\left( 2t-1 \right)\left( t+1 \right)$
So, we get,
$\Rightarrow \left( 2t-1 \right)\left( t+1 \right)=0$
So, we get,
$\Rightarrow \left( 2t-1 \right)=0;\left( t+1 \right)=0$
So, we get,
$\Rightarrow t=\dfrac{1}{2};t=-1$
As t = cos x, so we get,
$\cos x=\dfrac{1}{2};\cos x=-1$
For $\cos x=\dfrac{1}{2},$
As cos x is positive in quadrant I and IV, so $\cos x=\dfrac{1}{2}$ for x lie in II and IV quadrants. So, as we know $\cos {{60}^{\circ }}=\dfrac{1}{2},$ so in quadrant first $\cos x=\dfrac{1}{2}$ for $x=\dfrac{\pi }{3}.$
In quadrant IV, the solution for $\cos x=\dfrac{1}{2}$ is given as $x=2\pi -\theta$ where $\theta$ is the solution for $\cos x=\dfrac{1}{2}$ in quadrant I. So, as the solution for $\cos x=\dfrac{1}{2}$ in quadrant I is $\dfrac{\pi }{3}.$ So, we get,
$x=2\pi -\dfrac{\pi }{3}=\dfrac{5\pi }{3}$
For cos x = – 1,
We know cos x = – 1 is only for $\pi .$ So, cos x = – 1 for $x=\pi .$ Hence, we get the solution for $2{{\sin }^{2}}x-\cos x=1$ are $\dfrac{\pi }{3},\dfrac{5\pi }{3},\pi .$

Note:
Since sin and cos are periodic functions, so remember the degree of these functions will not determine the number of solutions. The degree of polynomial only defines the number of possible solutions as cos x = 1 will have infinitely many solutions but its degree is just 1. Also, remember that we need to have good knowledge about the sign of different ratios in different quadrants.