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erik [133]
2 years ago
5

Solve 4 cos2x-3 = 0 for all real values of x.

Mathematics
1 answer:
vichka [17]2 years ago
3 0

4 cos² x - 3 = 0

4 cos² x = 3

cos² x = 3/4

cos x = ±(√3)/2

Fixing the squared cosine doesn't discriminate among quadrants.  There's one in every quadrant

cos x = ± cos(π/6)

Let's do plus first.  In general, cos x = cos a has solutions x = ±a + 2πk integer k

cos x = cos(π/6)

x = ±π/6 + 2πk

Minus next.

cos x = -cos(π/6)

cos x = cos(π - π/6)

cos x = cos(5π/6)

x = ±5π/6 + 2πk

We'll write all our solutions as

x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k

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\mathbb{ \ \underline{ \:  \:  ANSWER  : }}

1. Find the percent markup.

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\mathbb{ \ \underline{ \:  \:  SOLUTION    : }}

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