Find general solution for secx = 2
Answer
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Hint: First we will convert sec into cos and then write that for what value of cos of the angle we get $\dfrac{1}{2}$, and then we will use the general solution of cos to find all the possible solutions, and we can see that there will be infinitely many solutions of x for which it gives $\cos x=\dfrac{1}{2}$.
Complete step-by-step answer:
Let’s convert sec into cos using the formula $\cos x=\dfrac{1}{\sec x}$
Hence, for secx = 2 we get $\cos x=\dfrac{1}{2}$.
Let’s first find the value of angle for which we get $\dfrac{1}{2}$.
Now we need to find that at which quadrant cos is positive,
We know that cos is positive in ${4}^{th}$ and ${1}^{st}$ quadrant, so $\dfrac{\pi }{3}$ and $\dfrac{-\pi }{3}$ both are the correct value,
Here, we will take $\dfrac{\pi }{3}$.
Now we know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$
Hence, we get $\cos x=\cos \dfrac{\pi }{3}$
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for $\cos x=\cos \dfrac{\pi }{3}$ we get,
$x=2n\pi \pm \dfrac{\pi }{3}$
Here n = integer.
Hence, from this we can see that we will get infinitely many solutions for x as we change the value of n.
Note: The formula for finding the general solution of cos is very important and must be kept in mind. In the above solution we have taken the value of $\alpha $ we have taken was$\dfrac{\pi }{3}$ , but one can also take the value of $\alpha $ as $\dfrac{-\pi }{3}$ , as it lies in the ${4}^{th}$ quadrant and gives positive value for cos. And then one can use the same formula for the general solution and replace the value of $\alpha $ with $\dfrac{-\pi }{3}$ to get the answer, which is also correct.
Complete step-by-step answer:
Let’s convert sec into cos using the formula $\cos x=\dfrac{1}{\sec x}$
Hence, for secx = 2 we get $\cos x=\dfrac{1}{2}$.
Let’s first find the value of angle for which we get $\dfrac{1}{2}$.
Now we need to find that at which quadrant cos is positive,
We know that cos is positive in ${4}^{th}$ and ${1}^{st}$ quadrant, so $\dfrac{\pi }{3}$ and $\dfrac{-\pi }{3}$ both are the correct value,
Here, we will take $\dfrac{\pi }{3}$.
Now we know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$
Hence, we get $\cos x=\cos \dfrac{\pi }{3}$
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha $ then the general solution is:
$\theta =2n\pi \pm \alpha $
Now using the above formula for $\cos x=\cos \dfrac{\pi }{3}$ we get,
$x=2n\pi \pm \dfrac{\pi }{3}$
Here n = integer.
Hence, from this we can see that we will get infinitely many solutions for x as we change the value of n.
Note: The formula for finding the general solution of cos is very important and must be kept in mind. In the above solution we have taken the value of $\alpha $ we have taken was$\dfrac{\pi }{3}$ , but one can also take the value of $\alpha $ as $\dfrac{-\pi }{3}$ , as it lies in the ${4}^{th}$ quadrant and gives positive value for cos. And then one can use the same formula for the general solution and replace the value of $\alpha $ with $\dfrac{-\pi }{3}$ to get the answer, which is also correct.
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