Question

How do you solve ${{\csc }^{2}}x-\csc x-2=0$ between the interval $0\underline{ < }x\underline{ < }2\pi$?

Answer 1

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Trigonometric identities are the equations involving the trigonometric functions that are true for every value of the variables involved. These identities are true for right angled triangles. So, the Pythagorean identity of sine function is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step by step answer:
As per the given question, we need to solve the given trigonometric expression using trigonometric identities and algebraic formulae. Here, we are given the expression ${{\csc }^{2}}x-\csc x-2=0$
In the given expression, if we assume then the expression becomes
${{a}^{2}}-a-2=0$
The equation looks like a quadratic polynomial. So, we can use factoring by grouping method to solve the above equation.
The coefficient of ${{a}^{2}}$ and constant terms are of the opposite sign and their product is -2. That is, we have a and c coefficients with the opposite sign in the polynomial of the form $a{{x}^{2}}+bx+c$.
Hence, we would split -1, which is the coefficient of x, into two parts, whose sum is -1 and product is -2. These are 1 and -2.
So, we write it as

& \Rightarrow {{a}^{2}}-a-2=0 \\\ & \Rightarrow {{a}^{2}}-2a+a-2=0 \\\ & \Rightarrow a\left( a-2 \right)+1\left( a-2 \right)=0 \\\ & \Rightarrow \left( a+1 \right)\left( a-2 \right)=0 \\\ \end{aligned}$$ Now we equate the equation then we get the values of a as 2,-1. Now we substitute $$\csc x$$ in a then we get the values as $$\csc x=2,\csc x=-1$$. We know that the reciprocal of $$\sin x$$ is $$\csc x$$. On substituting this $$\sin x=\dfrac{1}{2},\sin x=-1$$. We know that $$\sin x=-1$$ when $$x=\dfrac{3\pi }{2}$$ in the interval $$0\underline{ < }x\underline{ < }2\pi $$ and $$\sin x=\dfrac{1}{2}$$ when $$x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$$ in the interval $$0\underline{ < }x\underline{ < }2\pi $$. **Therefore, the equation $${{\csc }^{2}}x-\csc x-2=0$$ satisfies $$x=\dfrac{\pi }{6},\dfrac{3\pi }{2},\dfrac{5\pi }{6}$$ in the interval $$0\underline{ < }x\underline{ < }2\pi $$.** **Note:** In order to solve such types of questions, we need to have enough knowledge over trigonometric functions and identities. We also need to know the algebraic formulae to simplify the expressions. We must avoid calculation mistakes to get the expected answers.