Algebraically solving sin and cos functions, helpl please!

**i cant find any symbols on this site so if its confusing please do tell. **

sin[2(x-(pi/4))]=sq root2/2 for 0< x <2pi and for all feta (x=feta)

my attempt:

2(x-(pi/4))= pi/4 , 7pi/4

x-(pi/4) =pi/8 , 7pi/8

x = 3pi/8 9pi/8

based on the domain , only 3pi/8 fits.

for all feta = 3pi/8 +pi(n) , 9pi/8 + pi(n) where n is an element of the integers.

**4cos(x+30degrees)+5=7 for 0<x<360 and for all feta (x=feta)**

COS(X+30)=1/2

x+30 = 1.047197551

i dont know where to go from here

**2sin[2(x-(pi/6))]=-sqroot3 for 0<x<2pi and for all feta (x=feta) **

sin[2(x-(pi/6))] = -sqroot3/2

2(x-(pi/6))= pi/3 5pi/3

x-(pi/6) = pi/6 ,5pi/6

x= pi/3 and pi (both work)

for all: pi +pi(N) , pi/3 + pi(n) where n is an element of the integers.

3sin(x-1)=2 0<x<360 round to the tenth of a degree and for all feta (x=feta)

sin(x-1)=2/3

x-1=.7297276567

again im having problems with these decimal and degrees ones

and if you can check over the two i finished please and thank you , if i made mistakes please point them out even the obvious ones

i have a quiz tomorrow so if can someone PLEASE help me out thank you!

Re: Algebraically solving sin and cos functions, helpl please!

Quote:

Originally Posted by

**skg94** **i cant find any symbols on this site so if its confusing please do tell. **

sin[2(x-(pi/4))]=sq root2/2 for 0< x <2pi and for all feta (x=feta)

my attempt:

2(x-(pi/4))= pi/4 , 7pi/4

x-(pi/4) =pi/8 , 7pi/8

x = 3pi/8 9pi/8

based on the domain , only 3pi/8 fits.

for all feta = 3pi/8 +pi(n) , 9pi/8 + pi(n) where n is an element of the integers.

**4cos(x+30degrees)+5=7 for 0<x<360 and for all feta (x=feta)**

COS(X+30)=1/2

x+30 = 1.047197551

i dont know where to go from here

**2sin[2(x-(pi/6))]=-sqroot3 for 0<x<2pi and for all feta (x=feta) **

sin[2(x-(pi/6))] = -sqroot3/2

2(x-(pi/6))= pi/3 5pi/3

x-(pi/6) = pi/6 ,5pi/6

x= pi/3 and pi (both work)

for all: pi +pi(N) , pi/3 + pi(n) where n is an element of the integers.

3sin(x-1)=2 0<x<360 round to the tenth of a degree and for all feta (x=feta)

sin(x-1)=2/3

x-1=.7297276567

again im having problems with these decimal and degrees ones

and if you can check over the two i finished please and thank you , if i made mistakes please point them out even the obvious ones

i have a quiz tomorrow so if can someone PLEASE help me out thank you!

Re: Algebraically solving sin and cos functions, helpl please!

well thank you for that, could you potentially help me with the others? please that helped me alot

how do i know if my second angle should equal to pi or 2pi? based on the period? at the start

where you did 3pi/4 and not 7pi/4

Re: Algebraically solving sin and cos functions, helpl please!

Re: Algebraically solving sin and cos functions, helpl please!

i just have difficulty with the ref and principal angles that is my problem i believe so if i was taught correctly

q=ref=prince

q2 ref+prince=pi

q3=prince-ref=pi

q4=ref+prince=2pi

for the second one

1.04..-30 = -28.95 degrees? saying that its in the first quadrant or the 4th since its negative, is that my reference angle? then i find my princaple angle for the first quadrant?

i now understand the positive and negative sides, for the third one since its negative root3/2 it would make it the 3rd and 4th quadrants so pi/3 and 4pi/3 correct?

can you please help me with these decimal ones? if you can

thanks for everything so far

Re: Algebraically solving sin and cos functions, helpl please!

Quote:

Originally Posted by

**skg94** i just have difficulty with the ref and principal angles that is my problem i believe so if i was taught correctly

q=ref=prince

q2 ref+prince=pi

q3=prince-ref=pi

q4=ref+prince=2pi

for the second one

1.04..-30 = -28.95 degrees? saying that its in the first quadrant or the 4th since its negative, is that my reference angle? then i find my princaple angle for the first quadrant?

i now understand the positive and negative sides, for the third one since its negative root3/2 it would make it the 3rd and 4th quadrants so pi/3 and 4pi/3 correct?

can you please help me with these decimal ones? if you can

thanks for everything so far

It's actually:

Quadrant 1: Angle = Reference Angle.

Quadrant 2: Angle = Pi - Reference Angle

Quadrant 3: Angle = Pi + Reference Angle

Quadrant 4: Angle = 2Pi - Reference Angle.

Re: Algebraically solving sin and cos functions, helpl please!

so do i have my -28.95 correct?

so if that is in q4, to find my angle in q1, i do

2pi-(-28.95)?