
How do you solve $2+\sec x=0$ and find all solutions in the interval ${{0}^{\circ }}\le x<{{360}^{\circ }}$ ?
Answer
543.3k+ views
Hint: In order to find solution to this trigonometric equation, we will use trigonometric functions and in order to find the solution in the interval of $x$ within the range, we will subtract $x$ from $360$, that is first by using trigonometric function properties we will find first value of $x$ and then we will subtract that value from $360$ to get another value of $x$.
Complete step by step answer:
In geometry, trigonometric functions are the functions that are used to find the unknown angle or side of a right-angled triangle.
There are three basic trigonometry functions: Sine, Cosine and Tangent.
The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry and the three functions which are cotangent, secant and cosecant can be derived from the primary functions.
So Basically, the other three functions are often used as compare to the primary trigonometric functions.
Now coming to our question, we have:
$\Rightarrow 2+\sec x=0$
Now, we will use trigonometric function properties, that is $\sec x=\dfrac{1}{\cos x}$ :
Therefore, on implying and shifting $2$ on the other side, we get our equation as:
$\Rightarrow \dfrac{1}{\cos x}=-2$
On re-rearranging the terms in the expression, we get:
$\Rightarrow \cos x=-\dfrac{1}{2}$
Now, on taking cosine on right-hand side, we get:
$\Rightarrow x={{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$
That is
$\Rightarrow x={{120}^{\circ }}$ since ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)={{120}^{\circ }}$
Now in order to find the other value of $x$ within the range, we will subtract $x$ from $360$.
Therefore, we get:
$\Rightarrow x=360-120$
On simplifying, we get:
$\Rightarrow x={{240}^{\circ }}$
Therefore, the solutions in the interval ${{0}^{\circ }}\le x<{{360}^{\circ }}$ is:
$\Rightarrow x={{120}^{\circ }},{{240}^{\circ }}$, which is the required answer.
Note: Trigonometric functions are also known as a Circular Functions can be simply defined as the functions of an angle of a triangle.
It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions.
The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.
Complete step by step answer:
In geometry, trigonometric functions are the functions that are used to find the unknown angle or side of a right-angled triangle.
There are three basic trigonometry functions: Sine, Cosine and Tangent.
The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry and the three functions which are cotangent, secant and cosecant can be derived from the primary functions.
So Basically, the other three functions are often used as compare to the primary trigonometric functions.
Now coming to our question, we have:
$\Rightarrow 2+\sec x=0$
Now, we will use trigonometric function properties, that is $\sec x=\dfrac{1}{\cos x}$ :
Therefore, on implying and shifting $2$ on the other side, we get our equation as:
$\Rightarrow \dfrac{1}{\cos x}=-2$
On re-rearranging the terms in the expression, we get:
$\Rightarrow \cos x=-\dfrac{1}{2}$
Now, on taking cosine on right-hand side, we get:
$\Rightarrow x={{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$
That is
$\Rightarrow x={{120}^{\circ }}$ since ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)={{120}^{\circ }}$
Now in order to find the other value of $x$ within the range, we will subtract $x$ from $360$.
Therefore, we get:
$\Rightarrow x=360-120$
On simplifying, we get:
$\Rightarrow x={{240}^{\circ }}$
Therefore, the solutions in the interval ${{0}^{\circ }}\le x<{{360}^{\circ }}$ is:
$\Rightarrow x={{120}^{\circ }},{{240}^{\circ }}$, which is the required answer.
Note: Trigonometric functions are also known as a Circular Functions can be simply defined as the functions of an angle of a triangle.
It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions.
The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.
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