How do you solve $\cos e{c^2}x - 2 = 0$ and find all the solutions in the interval $[0,2\pi )$ ?
Answer
631.2k+ views
Hint: In this question, we are given a trigonometric equation and we have been asked to find the value of x in the given interval. Convert the given trigonometric ratio in terms of sin. Then shift all the constants to RHS and then, square root both the sides. You will get the value of the ratio. Find the angles at which the answer is that value within the given interval.
Formula used: $\sin x = \dfrac{1}{{\cos ecx}}$
Complete step-by-step solution:
We have been asked to solve $\cos e{c^2}x - 2 = 0$.
First, convert cosec into sin.
$ \Rightarrow \cos e{c^2}x - 2 = 0$ …. (given)
Using the formula and we get$\sin x = \dfrac{1}{{\cos ecx}}$,
$ \Rightarrow \dfrac{1}{{{{\sin }^2}x}} - 2 = 0$
Shifting the constant terms to the other side,
$ \Rightarrow \dfrac{1}{{{{\sin }^2}x}} = 2$
On rewriting this we get
$ \Rightarrow \dfrac{1}{2} = {\sin ^2}x$
Square rooting both the sides,
$ \Rightarrow \sqrt {\dfrac{1}{2}} = \sqrt {{{\sin }^2}x} $
On simplifying, we get,
$ \Rightarrow \pm \dfrac{1}{{\sqrt 2 }} = \sin x$
Now, we have to find the angle in which the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$.
One such angle is $\dfrac{\pi }{4}$ in the first quadrant. In second quadrant, such angle will be $\dfrac{{3\pi }}{4}\left( { = \dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)$
There are more such angles in the third and fourth quadrant.
Let us find them out.
In third quadrant, the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$ at $\dfrac{{5\pi }}{4}\left( { = \pi + \dfrac{\pi }{4}} \right)$.
In fourth quadrant, the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$ at $\dfrac{{7\pi }}{4}\left( { = \dfrac{{3\pi }}{2} + \dfrac{\pi }{4}} \right)$.
Hence, the solution of $\cos e{c^2}x - 2 = 0$ in the interval $[0,2\pi )$ is $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$.
Note: Open and close brackets:
There are 2 types of brackets. The square brackets [-] are called close brackets. If these brackets are used, then the numbers are included in the range. For example: In $\left[ {1,4} \right]$, the numbers included are $1,2,3,4$.
Another type of bracket is round bracket (-). These are called open brackets. If these brackets are used, then the numbers are not included in the range. For example: In $\left( {1,4} \right)$, the numbers included are $2,3$.
Formula used: $\sin x = \dfrac{1}{{\cos ecx}}$
Complete step-by-step solution:
We have been asked to solve $\cos e{c^2}x - 2 = 0$.
First, convert cosec into sin.
$ \Rightarrow \cos e{c^2}x - 2 = 0$ …. (given)
Using the formula and we get$\sin x = \dfrac{1}{{\cos ecx}}$,
$ \Rightarrow \dfrac{1}{{{{\sin }^2}x}} - 2 = 0$
Shifting the constant terms to the other side,
$ \Rightarrow \dfrac{1}{{{{\sin }^2}x}} = 2$
On rewriting this we get
$ \Rightarrow \dfrac{1}{2} = {\sin ^2}x$
Square rooting both the sides,
$ \Rightarrow \sqrt {\dfrac{1}{2}} = \sqrt {{{\sin }^2}x} $
On simplifying, we get,
$ \Rightarrow \pm \dfrac{1}{{\sqrt 2 }} = \sin x$
Now, we have to find the angle in which the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$.
One such angle is $\dfrac{\pi }{4}$ in the first quadrant. In second quadrant, such angle will be $\dfrac{{3\pi }}{4}\left( { = \dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)$
There are more such angles in the third and fourth quadrant.
Let us find them out.
In third quadrant, the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$ at $\dfrac{{5\pi }}{4}\left( { = \pi + \dfrac{\pi }{4}} \right)$.
In fourth quadrant, the value of $\sin x$ is $ \pm \dfrac{1}{{\sqrt 2 }}$ at $\dfrac{{7\pi }}{4}\left( { = \dfrac{{3\pi }}{2} + \dfrac{\pi }{4}} \right)$.
Hence, the solution of $\cos e{c^2}x - 2 = 0$ in the interval $[0,2\pi )$ is $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$.
Note: Open and close brackets:
There are 2 types of brackets. The square brackets [-] are called close brackets. If these brackets are used, then the numbers are included in the range. For example: In $\left[ {1,4} \right]$, the numbers included are $1,2,3,4$.
Another type of bracket is round bracket (-). These are called open brackets. If these brackets are used, then the numbers are not included in the range. For example: In $\left( {1,4} \right)$, the numbers included are $2,3$.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

