Question

Solve the equation on the interval [0,2pi] 5sec(x)+7=-3

Answer 1

Answer:

Hello,

$x\in \bigg\{\dfrac{2\pi}{3} ,\dfrac{4\pi}{3}\bigg\}$

Step-by-step explanation:

$5*sec(x)+7=-3\\\\sec(x)=\dfrac{-10}{5} \\\\\dfrac{1}{cos(x)} =-2\\\\cos(x)=-\dfrac{1}{2} \\\\x=120^o\ or\ x=240^o$

Answer 2

Answer:

$\sf \boxed{\sf x = \dfrac{2\pi}{3},\dfrac{4\pi}{3}}$

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 ,

$\sf\longrightarrow 5 sec\ x +7 = -3$

Add -7 to both sides ,

$\sf\longrightarrow 5sec\ x = -10$

Divide both sides by 5 ,

$\sf\longrightarrow sec \ x =\dfrac{-10}{5}$

Simplify ,

$\sf\longrightarrow sec \ x =-2$

Now solve for x ,

$\sf\longrightarrow x = sec^{-1}(-2)$

Simplify ,

$\sf\longrightarrow x = \dfrac{2\pi}{3}$

The secant function is negative in the second and third quadrants. Subtracting the reference angle from 2π to find the solution in the third quadrant to find the solution second solution.

$\sf\longrightarrow x = 2\pi -\dfrac{2\pi}{3}$

Simplify ,

$\sf\longrightarrow x = \dfrac{4\pi}{3}$

Now here the period of sec x is 2π . Therefore ,

$\sf\longrightarrow x = \dfrac{2\pi}{3} +2\pi n , \dfrac{4\pi}{3}+2\pi n , \textsf{ for any integer n } .$

Therefore all the possible solutions are ,

$\sf\longrightarrow \boxed{\blue{\sf x = \dfrac{2\pi}{3},\dfrac{4\pi}{3}}}$

Hence the required answer is 2π/3 and 4π/3.