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AleksAgata [21]
3 years ago
10

Solve the following equations for x,

Mathematics
1 answer:
stellarik [79]3 years ago
7 0

(i) 3 csc²(<em>x</em>) - 4 = 0

3 csc²(<em>x</em>) = 4

csc²(<em>x</em>) = 4/3

sin²(<em>x</em>) = 3/4

sin(<em>x</em>) = ± √3/2

<em>x</em> = arcsin(√3/2) + 2<em>nπ</em>  <u>or</u>   <em>x</em> = arcsin(-√3/2) + 2<em>nπ</em>

<em>x</em> = <em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + 2<em>nπ</em>

where <em>n</em> is any integer. The general result follows from the fact that sin(<em>x</em>) is 2<em>π</em>-periodic.

In the interval 0 ≤ <em>x</em> ≤ 2<em>π</em>, the first family of solutions gives <em>x</em> = <em>π</em>/3 and <em>x</em> = 4<em>π</em>/3 for <em>n</em> = 0 and <em>n</em> = 1, respectively; the second family gives <em>x</em> = 2<em>π</em>/3 and <em>x</em> = 5<em>π</em>/3 for <em>n</em> = 1 and <em>n</em> = 2.

(ii) 4 cos²(<em>x</em>) + 2 cos(<em>x</em>) - 2 = 0

2 cos²(<em>x</em>) + cos(<em>x</em>) - 1 = 0

(2 cos(<em>x</em>) - 1) (cos(<em>x</em>) + 1) = 0

2 cos(<em>x</em>) - 1 = 0   <u>or</u>   cos(<em>x</em>) + 1 = 0

2 cos(<em>x</em>) = 1   <u>or</u>   cos(<em>x</em>) = -1

cos(<em>x</em>) = 1/2   <u>or</u>   cos(<em>x</em>) = -1

[<em>x</em> = arccos(1/2) + 2<em>nπ</em>   <u>or</u>   <em>x</em> = 2<em>π</em> - arccos(1/2) + 2<em>nπ</em>]   <u>or</u>   <em>x</em> = arccos(-1) + 2<em>nπ</em>

[<em>x</em> = <em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   <em>x</em> = 5<em>π</em>/3 + 2<em>nπ</em>]   <u>or</u>   <em>x</em> = <em>π</em> + 2<em>nπ</em>

For 0 ≤ <em>x</em> ≤ 2<em>π</em>, the solutions are <em>x</em> = <em>π</em>/3, <em>x</em> = 5<em>π</em>/3, and <em>x</em> = <em>π</em>.

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