
How do you solve \[\sec x.\csc x - 2\csc x = 0\]?
Answer
546.3k+ views
Hint:We can solve this equation by using reciprocals of trigonometric functions. We know the reciprocals of sine, cosine, and tangent functions are cosecant, secant and cotangent
respectively.
We use this to simplify the trigonometric equation. Here we need to solve the trigonometric equation for ‘x’.
Complete step by step solution:
Given, \[\sec x.\csc x - 2\csc x = 0\].
Adding \[2\csc x\]on both sides of a trigonometric equation we have,
\[ \Rightarrow \sec x.\csc x = 2\csc x\]
Now divide the whole equation with \[\csc x\],
\[ \Rightarrow \dfrac{{\sec x.\csc x}}{{\csc x}} = \dfrac{{2\csc x}}{{\csc x}}\]
Cancelling we have,
\[ \Rightarrow \sec x = 2\]
To solve this easily, let’s take the reciprocal of whole equation
\[ \Rightarrow \dfrac{1}{{\sec x}} = \dfrac{1}{2}\]
We know that the reciprocal of secant is cosine.
\[ \Rightarrow \cos x = \dfrac{1}{2}\]
We know that \[\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\], then above becomes
\[ \Rightarrow \cos x = \cos \left( {\dfrac{\pi }{3}} \right)\]
We get,
\[ \Rightarrow x = \dfrac{\pi }{3}\] is the required answer.
Additional information: Trigonometry comes from the two roots, trigonon (or “triangle”) and metria (or “measure”). The study of trigonometry is thus the study of measurements of triangles. The first objects that come to mind may be the lengths of the sides, the angles of the triangle, or the area contained in the triangle. Using the angles and sides of a triangle we have defined all the six trigonometric functions.
Note: We know the reciprocals of sine, cosine, and tangent functions are cosecant, secant and cotangent respectively. Also cosecant, secant and cotangent functions are sine, cosine, and tangent respectively. In the above problem we have \[\sec x = 2\]. If we remember the standard
trigonometric value at a standard angle we know \[\sec \left( {\dfrac{\pi }{3}} \right) = 2\]. Then we
will have \[ \Rightarrow \sec x = \sec \left( {\dfrac{\pi }{3}} \right)\]. Thus we have \[ \Rightarrow x = \dfrac{\pi }{3}\]. In both cases we will have the same answer.
respectively.
We use this to simplify the trigonometric equation. Here we need to solve the trigonometric equation for ‘x’.
Complete step by step solution:
Given, \[\sec x.\csc x - 2\csc x = 0\].
Adding \[2\csc x\]on both sides of a trigonometric equation we have,
\[ \Rightarrow \sec x.\csc x = 2\csc x\]
Now divide the whole equation with \[\csc x\],
\[ \Rightarrow \dfrac{{\sec x.\csc x}}{{\csc x}} = \dfrac{{2\csc x}}{{\csc x}}\]
Cancelling we have,
\[ \Rightarrow \sec x = 2\]
To solve this easily, let’s take the reciprocal of whole equation
\[ \Rightarrow \dfrac{1}{{\sec x}} = \dfrac{1}{2}\]
We know that the reciprocal of secant is cosine.
\[ \Rightarrow \cos x = \dfrac{1}{2}\]
We know that \[\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\], then above becomes
\[ \Rightarrow \cos x = \cos \left( {\dfrac{\pi }{3}} \right)\]
We get,
\[ \Rightarrow x = \dfrac{\pi }{3}\] is the required answer.
Additional information: Trigonometry comes from the two roots, trigonon (or “triangle”) and metria (or “measure”). The study of trigonometry is thus the study of measurements of triangles. The first objects that come to mind may be the lengths of the sides, the angles of the triangle, or the area contained in the triangle. Using the angles and sides of a triangle we have defined all the six trigonometric functions.
Note: We know the reciprocals of sine, cosine, and tangent functions are cosecant, secant and cotangent respectively. Also cosecant, secant and cotangent functions are sine, cosine, and tangent respectively. In the above problem we have \[\sec x = 2\]. If we remember the standard
trigonometric value at a standard angle we know \[\sec \left( {\dfrac{\pi }{3}} \right) = 2\]. Then we
will have \[ \Rightarrow \sec x = \sec \left( {\dfrac{\pi }{3}} \right)\]. Thus we have \[ \Rightarrow x = \dfrac{\pi }{3}\]. In both cases we will have the same answer.
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