Prelim Trig (1 Viewer)

Marc26

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hey guys can you show me a step by step process of these, they are a bit tricky at first..
Solve for 0 < theta < 360 to the nearest minute:
WolframAlpha--solve_8cos2x__2sinx__7--2012-05-30_0239.gif
 
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8cos^2(x) = 2sin(x) +7

8(1-sin^2(x)) = 2sin(x) +7

8 -8sin^2(x) = 2sin(x) +7

Move everything to the RHS

8sin^2(x) +2sin(x) -1 =0

Two numbers that add to +2 and multiply to give (8 x -1) = -8

4 & -2

8sin^2(x) +4sin(x) -2sin(x) -1 =0
4sin(x) [ 2sin(x) +1] -(2sin(x) +1) =0

(2sin(x)+1) [ 4sin(x) -1] =0

Therefor

2sin(x) +1 =0 or 4sin(x) -1 =0

----

2sin(x) +1 =0 ---> sin(x) = -1/2

Related angle = sin^(-1) (1/2) = 30degrees

Sin is negative in 3rd and 4th quadrants

Therefore x= (180+30), (360-30) = 210, 330


--------------

4sin(x) -1 =0

sin(x) = 1/4

Related angle = sin^(-1) (1/4) = 14.4775.... degrees

Sin is positive in first and second quadrants

Therefore, x= 14.4775 , (180-14.4775) = 14.4775 , 165.5225 degrees

Changing to degrees and minutes

x= 14*29' , 165*31'

----

Therefore all solutions are x= 14*29' , 165*31' , 210*, 330*
 

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