
Gaukhar M.
asked 10/30/20please help with these
a) Solve
8cos(2t)=8sin2(t)+3 for all solutions
0≤t<2π
t =
b) Solve 8sin(2w)+11sin(w)=0 for all solutions
0≤w<2π
c) Solve
2sin2(t)+3sin(t)+1=0 for all solutions.
Your initial angle should be between 0 and 2π (not including 2π , and your initial angles should go from least to greatest. Consider 2nπ to have an initial angle of 0). Use n as your multiplier.
1 Expert Answer
Need to use your double angle formula to get 8( cos^2(t) -sin^2(t)= 8 sin^2(t) +3
8cos^2(t) -8sin^2(t) =8sin^2(t) +3 Use the identity cos^2(t) = 1-sin^2(t)
8(1-sin^2(t)) -8sin^2(t) =8sin^2(t) +3\
8- 8sin^2(t) -8sin^2(t) = 8sin^2(t) +3
8 -16sin^2(t) =8sin^2(t) +3
24 sin^2(t) = 5 sin^2(t) =5/24 sin(t) = +/- sqroot (5/24) Take the inverse sin to get the angles +/- 27.16 degrees You will have angles in all 4 quadrants 27.16, (180-27.16) 180+27.16 and -27.16
For the second problem use your double angle formula to change 8sin(2w) to 8( 2sin(w)cos(w) = 16sin(w)cos(w). I will leave it to you to solve from here
For problem 3 Factor the expression to (2sinx +1)(sinx+1) =0 so sinx = -1/2 and -1 I will leave it to you to find these angles
Gaukhar M.
Thank you so much for your help and time!appreciate it!11/01/20
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Mark M.
In a is it sin^2 t or sin 2t?10/30/20