Solving Trigonometric Equations...3

[nycmathguy]

Section 5.3

For 22, why is the solution pi•k, where k is any integer? Allow me to be honest. I am not really understanding this section.

 

[MathLover1]

18.
2-4sin^2(x)=0
2=4sin^2(x)
sin^2(x)=2/4
sin^2(x)=1/2
sin(x)=sqrt(1/2) or sin(x)=-sqrt(1/2)

if sin(x)=sqrt(1/2) =>x=π/4+2π*k, x= 3π/4+2π*k
if sin(x)=-sqrt(1/2)=>x=5π/4+2π*k, x=7π/4+2π*k

all solutions:
x= π/+2π*k
x=3π/4+2π*k
x=5π/4+2π*k
x=7π/4+2π*k

22.

sec ^2 (x)-1=0
(sec (x)-1)(sec (x)+1)=0

if sec (x)-1=0 =>sec (x)=1=>x=2pi*k
if sec (x)+1=0 =>sec (x)=-1=>x=pi +2pi*k
solutions:
x=2π*k
x=π+2π*k

why is the solution 2π•k, where k is any integer: because sec has period of 2π

reminder:

Period of y=sin x is 2π
Period of y=cos x is 2π
Period of y=tan x is π
Period of y=cot x is π
Period of y=sec x is 2π
Period of y=csc x is 2π

 

[nycmathguy]

18.
2-4sin^2(x)=0
2=4sin^2(x)
sin^2(x)=2/4
sin^2(x)=1/2
sin(x)=sqrt(1/2) or sin(x)=-sqrt(1/2)

if sin(x)=sqrt(1/2) =>x=π/4+2π*k, x= 3π/4+2π*k
if sin(x)=-sqrt(1/2)=>x=5π/4+2π*k, x=7π/4+2π*k

all solutions:
x= π/+2π*k
x=3π/4+2π*k
x=5π/4+2π*k
x=7π/4+2π*k

22.

sec ^2 (x)-1=0
(sec (x)-1)(sec (x)+1)=0

if sec (x)-1=0 =>sec (x)=1=>x=2pi*k
if sec (x)+1=0 =>sec (x)=-1=>x=pi +2pi*k
solutions:
x=2π*k
x=π+2π*k

why is the solution 2π•k, where k is any integer: because sec has period of 2π

reminder:

Period of y=sin x is 2π
Period of y=cos x is 2π
Period of y=tan x is π
Period of y=cot x is π
Period of y=sec x is 2π
Period of y=csc x is 2π

For 18, explain how you got the solutions posted.

For 22:

x = 0 + 2π*k, which of course becomes 2π*k.
x =π + 2π*k

 

[nycmathguy]

18.
2-4sin^2(x)=0
2=4sin^2(x)
sin^2(x)=2/4
sin^2(x)=1/2
sin(x)=sqrt(1/2) or sin(x)=-sqrt(1/2)

if sin(x)=sqrt(1/2) =>x=π/4+2π*k, x= 3π/4+2π*k
if sin(x)=-sqrt(1/2)=>x=5π/4+2π*k, x=7π/4+2π*k

all solutions:
x= π/+2π*k
x=3π/4+2π*k
x=5π/4+2π*k
x=7π/4+2π*k

22.

sec ^2 (x)-1=0
(sec (x)-1)(sec (x)+1)=0

if sec (x)-1=0 =>sec (x)=1=>x=2pi*k
if sec (x)+1=0 =>sec (x)=-1=>x=pi +2pi*k
solutions:
x=2π*k
x=π+2π*k

why is the solution 2π•k, where k is any integer: because sec has period of 2π

reminder:

Period of y=sin x is 2π
Period of y=cos x is 2π
Period of y=tan x is π
Period of y=cot x is π
Period of y=sec x is 2π
Period of y=csc x is 2π

Wonderful.

How did I do at the link below?

Solving Trigonometric Equations...4