Find, to the nearest hundredth place, the solutions to the following equation(s), where 0 ≤ x ≤ 360 degrees. `(sec(x)-1)(2sec(x)+3)=3` Thank you to everyone that helps.

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lemjay | High School Teacher | (Level 3) Senior Educator

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`(secx -1 )(2sec x + 3 ) = 3`

To solve this, first, multiply the right side. To do so, apply FOIL.

`2sec^2x +3secx - 2secx - 3 = 3`

`2sec^2x + secx - 3 = 3`

Then, set one side equal to zero. So, subtract both sides by 3.

`2sec^2x + secx - 3 - 3 =3 - 3`

`2sec^2x + secx - 6 =0`

Then, factor the left side.

`(2sec x -3)(secx + 2) = 0`

Set each factor equal to zero and solve for x.

For the first factor:

`2secx - 3 = 0`

`2secx = 3`

`secx = 3/2`

`1/cosx=3/2`

` `

`cosx = 2/3`

`x=cos^(-1)(-2/3)`

`x=48.19^o`

Since cosine is positive, the other value of the angle x lies on the fourth quadrant. Hence, the other angle is:

`x = 311.81^o`

For the second factor:

`secx + 2 = 0`

`secx = -2`

`1/cosx=-2`

`cosx = -1/2`

`x=cos^(-1)(-1/2)`

`x=120^o`

Since cosine is negative, the other angle x lies at the third quadrant. So, the other angle is:

`x = 240^o`

Therefore, the solution to the given equation in the given interval are `x=48.19^o, 120^o, 240^o,` and `311.81^o` .

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