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Leno4ka [110]
4 years ago
7

Which are solutions to 3sin^2 θ - 2sin θ = 0, 90°≤θ≤90°

Mathematics
1 answer:
jeka57 [31]4 years ago
4 0

Answer:

The solutions of the equation are -41.8° , 0.0° , 41.8°

Step-by-step explanation:

∵ 3 sin²Ф - 2 sinФ = 0 is a quadratic function

∵ The domain of the function is -90° ≤ Ф ≤ 90°

- Lets take sinФ as a common factor

∴ sinФ (3 sinФ - 2) = 0

- Two terms multiplied by each other = 0

∴ One of them must be equal zero

∴ sinФ = 0 or 3 sinФ - 2 = 0

- If sinФ = 0 ⇒ Ф = 0.0°

- If 3 sinФ - 2 = 0 ⇒ 3 sinФ = 2 ⇒ sin Ф = 2/3

∵ sinФ 2/3

∴ Ф = 41.8° ⇒ Ф ≤ 90° (in first quadrant)

∵ -90° ≤ Ф ≤ 90°

∴ Ф = -41.8° ⇒ Ф ≥ -90° (in fourth quadrant)

∴ The solutions of the equation are -41.8° , 0.0° , 41.8°

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<h3>Reasons </h3>

i.  Cube of any odd number is even. This statement is false because the cube of odd numbers are also odd

ii.  A perfect cube does not end with two zeros. This statement is True because the cube of a two digit number cannot be a 3 digit number and therefore cannot end with two zeros.

iii.  If square of a number ends with 5, then its cube ends with 25. This statement is False

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35^3= 35 x 35 x 35= 42875 which ends with 75

iv.  There is no perfect cube which ends with 8. This statement is False.

Using the illustration, the cube of 2 is given thus

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vii. The cube of a single digit number may be a single digit number. This statement is True because the cube of a single digit number can be a single, two or three digit number.

Learn more about cube numbers here:

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2 years ago
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