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postnew [5]
3 years ago
5

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

Mathematics
2 answers:
Umnica [9.8K]3 years ago
7 0

Answer:

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

Step-by-step 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

QveST [7]3 years ago
6 0

Answer:

\sf \boxed{\sf x = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}}

Step-by-step 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 =\dfrac{4}{3}

Put squareroot on both sides ,

\sf\longrightarrow sec \ x =\sqrt{\dfrac{4}{3} }

Simplify ,

\sf\longrightarrow sec\ x =\pm \dfrac{2}{\sqrt3}

Multiply numerator and denominator by √3 ,

\sf\longrightarrow \bf sec(x) = \dfrac{ 2\sqrt3}{3},-\dfrac{2\sqrt3}{3}

Solve for x ,

\sf\longrightarrow x = \dfrac{\pi}{6} +2\pi n , \dfrac{11\pi}{6}+2\pi n , \textsf{ for any integer n } .

Therefore all the possible solutions are ,

\sf\longrightarrow \boxed{\blue{\sf x = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}}}

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