will you take a look at the fun that even Wolfram had in solving the equation the way you typed it ???
http://www.wolframalpha.com/input/?i=solve+sin2x%2B1%3D-2sinx
I have a feeling that you meant:
sin^2 x + 1 = -2sinx
then
sin^2 x + 2sinx + 1 = 0
(sinx + 1)^2 = 0
sinx +1 = 0
sinx = -1
and x = 270° or 3π/2
confirm the typing of the others before I attempt them
Trigonometry Questions
1. Solve the equation sin2x+1=-2sinx for 0≤x≤2π
2. Solve the equation 7sin2x-4sin2x/cosx=-1 where 0≤x≤2π
3. Determine the exact value of cos2(theta) when tan(theta)=3/4 and π<theta<3π/2
4. The average number of customers, c, at a 24 hour sandwich shop per hour is modeled roughly by the equation c(h) = -5cos( π/12h) + 12, with h=0 representing midnight. What time of the day is peak business?
3 answers
thanks! i forgot to include the exponent. I re-typed them here:
2. Solve the equation 7sin^2x-4sin2x/cosx=-1 where 0≤x≤2π
3. Determine the exact value of cos2(theta) when tan(theta)=3/4 and π<theta<3π/2
4. The average number of customers, c, at a 24 hour sandwich shop per hour is modeled roughly by the equation c(h) = -5cos( π/12h) + 12, with h=0 representing midnight. What time of the day is peak business?
2. Solve the equation 7sin^2x-4sin2x/cosx=-1 where 0≤x≤2π
3. Determine the exact value of cos2(theta) when tan(theta)=3/4 and π<theta<3π/2
4. The average number of customers, c, at a 24 hour sandwich shop per hour is modeled roughly by the equation c(h) = -5cos( π/12h) + 12, with h=0 representing midnight. What time of the day is peak business?
2. 7sin^2x - 4sin2x/cosx = -1
7sin^2 x - 4(2sinxcosx)/cosx = -1
7sin^2x - 8sinx + 1 = 0
(7sinx - 1)(sinx - 1) = 0
sinx = 1/7 or sinx = 1
if sinx = 1
x = 90° or π/2
if sinx = 1/7, then
x = aprr 8.2° or 171.8°
3. cos 2Ø
= cos^2 Ø - sin^2 Ø
given: tan Ø = 3/4 (I recognize the 3-4-5 right-angled triangle
so sinØ = 3/5 and cosØ = 4/5
then cos 2Ø
= cos^2 Ø - sin^ Ø
= 16/25 - 9/25 = 7/25
4. c(h) = -5cos(π/12 h) + 12
c ' (h) = 5(π/12)sin(π/12 h)
= 0 for a max of c(h)
sin(π/12 h) = 0
π/12 h = 0 or π
1/12 h = 1
h = 12
12 hours from midnight would bring you to 12:00 noon
so the peak business is at noon
7sin^2 x - 4(2sinxcosx)/cosx = -1
7sin^2x - 8sinx + 1 = 0
(7sinx - 1)(sinx - 1) = 0
sinx = 1/7 or sinx = 1
if sinx = 1
x = 90° or π/2
if sinx = 1/7, then
x = aprr 8.2° or 171.8°
3. cos 2Ø
= cos^2 Ø - sin^2 Ø
given: tan Ø = 3/4 (I recognize the 3-4-5 right-angled triangle
so sinØ = 3/5 and cosØ = 4/5
then cos 2Ø
= cos^2 Ø - sin^ Ø
= 16/25 - 9/25 = 7/25
4. c(h) = -5cos(π/12 h) + 12
c ' (h) = 5(π/12)sin(π/12 h)
= 0 for a max of c(h)
sin(π/12 h) = 0
π/12 h = 0 or π
1/12 h = 1
h = 12
12 hours from midnight would bring you to 12:00 noon
so the peak business is at noon