Question

Show that no value of secant x can satisfy the equation 6 sec^2 X - 5 sec X + 1=0

Answer 1

Answer:

The given equation is

$6sec^{2} x-5secx+1=0$

Let $secx = t$

The equation now becomes

$6t^{2} -5t+1=0$

or, $6t^{2} -3t-2t+1=0$

or, $3t(2t-1)-1(2t-1)=0$

or, $(3t-1)(2t-1)=0$

This gives, $t=\frac{1}{2} , \frac{1}{3}$

Thus, $secx=\frac{1}{2} or \frac{1}{3}$

But we know that the range of secx is (-∞, -1] ∪ [1, ∞)

Thus there are no vales of x for which secx is equal to 1/2 or 1/3

Therefore, there are no values of secantx that can satisfy the given equation

Answer 2

Answer:

Step-by-step explanation:

Our question is: Show that no value of secant x can satisfy the equation 6 sec^2 X - 5 sec X + 1=0.

We have the following equation:

-- >  6 * (sec(x))^2 - 5 * sec(x) + 1 = 0

In order to check the values of secant of x that satisfies the above equation we will use the method of substitution.

In this case we say that sec(x) = m.

Substituting in the expression we get something like:

-- >  6*m^2 -5*m + 1 = 0

Using the formula that is given in the picture bellow we can get the solutions of the equation, and the solutions are m = 1/2 or m = 1/3.

Then we can substitute again the values of m with the initial made substitution, and we have sec(x) = 1/2 or sec(x) = 1/3.

We know that the range of the function sec(x) is (-∞, -1] ∪ [1, ∞). Hence those solutions do not belong to its range.

Therefore we can say that the equation given does not have any value of sec(x) that satisfies it.

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