Triginometric eqns: 5 - 7sin x = 2cos^2 x on [0, 2pi]

[anna_sims]

Solving trigonometric equations

I need help on solving these problems within specific intervals. I am having trouble solving for x the most. Any help is again appreciated.

1) 5-7sin x = 2cos^2 x

x in the interval [ -360, 450 degrees ]

2) 2csc^2 x = 3cot^2 x – 1

x in the interval [ 0, 2pi ]

 

[arthur ohlsten]

5-sinx = 2 cos^2x
for -360<x<450 degrees

we willconvert to one trig function by the equality
cos^2x=1-sin^2x substitute

5-sinx =2[1-sin^2x] clearing the bracket and rearanging
1)
2sin^2x -sinx +3=0 to help me, but is not neccessary let sinx = z
2z^2-z+3=0 useing the quadratic equation
z=1 +/-[1-24]^1/2 all over 4
no real solution

on a xy coordinate system draw a line at y=5
then plot 5-sinx , which is a sin curve around the line y=5 and varies between y=4 and y=6

then plot 2 cos^2x on the same graph.
2 cos^2 x is a curve between y=0 and y=2

the curves never cross, then are never equal

Arthur

 

[arthur ohlsten]

2)
2 csc^2x=3cot^2x
for 0<x<2pi

to make it easier for me , but not neccessary convert to cos and sin functions
csc^2x= 1/sin^2x
cot^2x= cos^2x/sin^2x

2 /sin^2x =3cos^2x/ sin^2x
multiply both sides by sin^2x, but sin^2x can't =0 or x=0*,pi*, or 2pi*
2= 3 cos^2x
cosx=+/- [2/3]^1/2
cosx= +/- 1/3 sqrt6
x= 35.3* or 324.7* answer
Arthur