Question

Match the equations with their solutions over the interval [0, 2π].

Answer 1

Explanation:

A). cos(x)tan(x) =
`(1)/(2)`

cos(x)
`(sin(x))/(cos(x))` =
`(1)/(2)`

sin(x) =
`(1)/(2)`

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

Since sine is positive in 1st and 2nd quadrant.

Therefore, x =
`(\pi )/(6) , (5\pi )/(6)`

B). sec(x)cot(x) + 2 = 0

[tex]\frac{1}{cos(x)}\frac{cos(x)}{sin(x)}=(-2)[/tex]

`(1)/(sin(x))=-2`

`sin(x)=-(1)/(2)`

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

Since sine is negative in 3rd and 4th quadrant.

Therefore, x =
`(7\pi )/(6)` and
`(11\pi )/(6)`

C). sin(x)cot(x) +
`(1)/(√(2) )` = 0

`[sin(x)](cos(x))/(sin(x))=-(1)/(√(2))`

cosx = -
`(1)/(√(2))`

`x=cos^(-1)(-(1)/(√(2)))`

Since cosine is negative in 2nd and 3rd quadrant.

Therefore, x =
`(3\pi )/(4)` and
`(5\pi )/(4)`

D). csc(x)tan(x) - 2 = 0

`((1)/(sinx))(sin(x))/(cos(x))=2`

`(1)/(cosx)=2`

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

Since x is positive in 1st and 4th quadrant.

x =
`(\pi )/(3)` and
`(5\pi )/(3)`

Answer 2

I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs are:

cos(x)tan(x) – ½ = 0 → π/6 and 5π/6

cos(π/6)tan(π/6) – ½ = 0

cos(5π/6)tan(5π/6) – ½ = 0

sec(x)cot(x) + 2 = 0 → 7π/6 and 11π/6

sec(7π/6)cot(7π/6) + 2 = 0

sec(11π/6)cot(11π/6) + 2 = 0

sin(x)cot(x) + 1/sqrt2 = 0 → 3π/4 and 5π/4

sin(3π/4)cot(3π/4) + 1/sqrt2 = 0

sin(5π/4)cot(5π/4) + 1/sqrt2 = 0

csc(x)tan(x) – 2 = 0 → π/3 and 5π/3

csc(π/3)tan(π/3) – 2 = 0

csc(5π/3)tan(5π/3) – 2 = 0