Respuesta :
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
Answer:
[tex]\sf \boxed{\sf x = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}}[/tex]
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 ,
[tex]\sf\longrightarrow 3(sec x)^2-4=0 [/tex]
This can be written as ,
[tex]\sf\longrightarrow 3sec^2x - 4 = 0 [/tex]
Add 4 to both sides of the equation ,
[tex]\sf\longrightarrow 3sec^2x = 4 [/tex]
Divide both sides by 3 ,
[tex]\sf\longrightarrow sec^2x =\dfrac{4}{3} [/tex]
Put squareroot on both sides ,
[tex]\sf\longrightarrow sec \ x =\sqrt{\dfrac{4}{3} }[/tex]
Simplify ,
[tex]\sf\longrightarrow sec\ x =\pm \dfrac{2}{\sqrt3} [/tex]
Multiply numerator and denominator by √3 ,
[tex]\sf\longrightarrow \bf sec(x) = \dfrac{ 2\sqrt3}{3},-\dfrac{2\sqrt3}{3} [/tex]
Solve for x ,
[tex]\sf\longrightarrow x = \dfrac{\pi}{6} +2\pi n , \dfrac{11\pi}{6}+2\pi n , \textsf{ for any integer n } . [/tex]
Therefore all the possible solutions are ,
[tex]\sf\longrightarrow \boxed{\blue{\sf x = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}}} [/tex]
