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enot [183]
3 years ago
14

Solve the equation 2cos^2x + 3sinx = 3

Mathematics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

x = 2 π n_1 + π/2 for n_1 element Z

or x = 2 π n_2 + (5 π)/6 for n_2 element Z or x = 2 π n_3 + π/6 for n_3 element Z

Step-by-step explanation:

Solve for x:

2 cos^2(x) + 3 sin(x) = 3

Write 2 cos^2(x) + 3 sin(x) = 3 in terms of sin(x) using the identity cos^2(x) = 1 - sin^2(x):

-1 + 3 sin(x) - 2 sin^2(x) = 0

The left hand side factors into a product with three terms:

-(sin(x) - 1) (2 sin(x) - 1) = 0

Multiply both sides by -1:

(sin(x) - 1) (2 sin(x) - 1) = 0

Split into two equations:

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

Add 1 to both sides:

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

Take the inverse sine of both sides:

x = 2 π n_1 + π/2 for n_1 element Z

or 2 sin(x) - 1 = 0

Add 1 to both sides:

x = 2 π n_1 + π/2 for n_1 element Z

or 2 sin(x) = 1

Divide both sides by 2:

x = 2 π n_1 + π/2 for n_1 element Z

or sin(x) = 1/2

Take the inverse sine of both sides:

Answer: x = 2 π n_1 + π/2 for n_1 element Z

or x = 2 π n_2 + (5 π)/6 for n_2 element Z or x = 2 π n_3 + π/6 for n_3 element Z

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