Question

Solve on the interval [0,2pi] 3(sec x)^2 -4=0

Answer 1

pi/6, 5pi/6, 7pi/6, 11pi/6

Explanation:

3(sec x)^2 -4=0

Add 4 on both sides:

3(sec x)^2=4

Divide 3 on both sides:

(sec x)^2=4/3

Take square root of both sides:

sec x=plus/minus sqrt(4/3)

Reciprocal identity:

cos x=plus/minus sqrt(3/4)

Simplify the radical:

cos x=plus/minus sqrt(3)/2

So we are looking on the unit circle where the first coordinate of the order pair is either sqrt(3)/2 or -sqrt(3)/2.

This happens at pi/6, 5pi/6, 7pi/6, 11pi/6

Answer 2

`\sf \boxed{\sf x = (\pi)/(6),(5\pi)/(6),(7\pi)/(6),(11\pi)/(6)}`

Explanation:

A trigonometric equation is given to us , and we need to find the solutions of the equation within the interval [ 0,2π ]

The given equation is ,

`\sf\longrightarrow 3(sec x)^2-4=0`

This can be written as ,

`\sf\longrightarrow 3sec^2x - 4 = 0`

Add 4 to both sides of the equation ,

`\sf\longrightarrow 3sec^2x = 4`

Divide both sides by 3 ,

`\sf\longrightarrow sec^2x =(4)/(3)`

Put squareroot on both sides ,

`\sf\longrightarrow sec \ x =\sqrt{(4)/(3) }`

Simplify ,

`\sf\longrightarrow sec\ x =\pm (2)/(\sqrt3)`

Multiply numerator and denominator by √3 ,

`\sf\longrightarrow \bf sec(x) = ( 2\sqrt3)/(3),-(2\sqrt3)/(3)`

Solve for x ,

`\sf\longrightarrow x = (\pi)/(6) +2\pi n , (11\pi)/(6)+2\pi n , \textsf{ for any integer n } .`

Therefore all the possible solutions are ,

`\sf\longrightarrow \boxed{\blue{\sf x = (\pi)/(6),(5\pi)/(6),(7\pi)/(6),(11\pi)/(6)}}`