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Trigonometric Equations/ Quadratic Formula/ Not exact values?

+1 vote
1) 8sin^2(x)- 4cos(x)-3=0

2) 8cos^2(x)+4cos(x)-5=0

 

How do i solve.

Round answers to nearest tenth o degree
asked Jan 18, 2013 in TRIGONOMETRY by homeworkhelp Scholar

1 Answer

+1 vote

1) 8sin^2(x)- 4cos(x)-3=0

8sin2(x) - 4cos(x) - 3 = 0

Note: sin2A + cos2A = 1 than sin2A = 1 - cos2A

8(1 - cos2x) - 4cos(x) - 3 = 0

Simplify

8 - 8cos2x - 4cos(x) - 3 = 0

- 8cos2x - 4cos(x) - 3 + 8 = 0

- 8cos2x - 4cos(x) + 5 = 0

Multiply each side by negative one.

8cos2x + 4cos(x) - 5 = 0

Let  cos(x) = t

Than

8t2 + 4t + 5 = 0

Quadratic formula image

Conspire the equation ax2+bx+c = 0

So substitute a = 8, b = 4, and c = 4 above the formula

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But t = cos(x)

There fore

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answered Jan 18, 2013 by richardson Scholar

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