Question

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

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

A). cos(x)tan(x) = $\frac{1}{2}$

cos(x)$\frac{sin(x)}{cos(x)}$ = $\frac{1}{2}$

sin(x) = $\frac{1}{2}$

$x=sinx^{-1}(\frac{1}{2} )$

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

Therefore, x = $\frac{\pi }{6} , \frac{5\pi }{6}$

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

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

$\frac{1}{sin(x)}=-2$

$sin(x)=-\frac{1}{2}$

$x=sin^{-1}(-\frac{1}{2})$

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

Therefore, x = $\frac{7\pi }{6}$ and $\frac{11\pi }{6}$

C). sin(x)cot(x) + $\frac{1}{\sqrt{2} }$ = 0

$[sin(x)]\frac{cos(x)}{sin(x)}=-\frac{1}{\sqrt{2}}$

cosx = -$\frac{1}{\sqrt{2}}$

$x=cos^{-1}(-\frac{1}{\sqrt{2}})$

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

Therefore, x = $\frac{3\pi }{4}$ and $\frac{5\pi }{4}$

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

$(\frac{1}{sinx})\frac{sin(x)}{cos(x)}=2$

$\frac{1}{cosx}=2$

cos(x) = $\frac{1}{2}$

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

x = $\frac{\pi }{3}$ and $\frac{5\pi }{3}$