Question

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Answer 1

Answer:   $\bold{b)\quad \dfrac{2\pi}{3},\dfrac{4\pi}{3},0}$

Step-by-step explanation:

Note the following identities: tan² x = sec²x - 1

                                                 $\sec x=\dfrac{1}{\cos}$

tan² x + sec x = 1

(sec² x -1) + sec x = 1

sec² x + sec x - 2 = 0

(sec x + 2)(sec x - 1) = 0

sec x + 2 = 0     sec x - 1 = 0

sec x = -2          sec x = 1

$\dfrac{1}{\cos x}=-2\qquad \dfrac{1}{\cos x}=1\\\\\cos x=-\dfrac{1}{2}\qquad \cos x=1\\\\x=\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\qquad x=0$

Answer 2

Answer:

Option B

Step-by-step explanation:

We have tan^2( x ) + sec( x ) = 1.

First subtract 1 from either side, applying the identity tan^2( x ) = - 1 + sec^2( x ) = - 2 + sec( x ) + sec^2( x ) = 0

One key method that you should use to solve like problems, is to substitute a rather large value with a variable, such as say a. Let sec( x ) = a,

- 2 + a + a^2 = 0,

By the zero product property, a = 1, a = - 2. Substitute this value of a back into sec( x ) -

sec( x ) = 1, sec( x ) = - 2

From this we can create inequality( s ) as follows -

sec( x ) = 1, 0 ≤ x ≤ 2π, x = 0 and x = 2π

sec( x ) = - 2, 0 ≤ x ≤ 2π, x = 2π / 3 and x = 4π / 3

As you can see, x is the following solutions -

0, 2π, 2π / 3, 4π / 3

Eliminate 2π, and it should be that your solution is option B