genzmintz
Answered

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Trigonometric equations
4sin^2(theta) + 4 = 5

Sagot :

angle46089

Answer:

Θ = (π/6) + πn

Θ = (5π/6) + πn

Step-by-step explanation:

4sin²Θ + 4 = 5

            -4    -4

4sin²Θ = 1

÷4         ÷4

sin²Θ = (1/4)

√sin²Θ = √(1/4)

sinΘ = (1/2), (-1/2)

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Θ = arcsin (1/2)

Θ = (π/6)

to find the quadrant subtract π

Θ = π - (π/6)

Θ = (5π/6)

Find the period

2π / |b|

b = 1

2π/1 = 2π

The sin Θ function is 2π, so values will repeat 2π in both directions.

Θ = (π/6) + 2πn (n is the variable)

Θ = (5π/6) + 2πn

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sin Θ = (-1/2)

Θ = arcsin (-1/2)

Θ = (-π/6)

To find the second function add π

Θ = 2π + (π/6) + π

Θ = (7π/6)

Find the period

2π/|b|

2π/1

2π

(-π/6) + 2π

2π       6         π

----- ×  ----- -  -----

1           6        6

Θ = (11π/6) will repeat every 2π in both directions

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Θ = (π/6) + 2πn

Θ = (5π/6) + 2πn

Θ = (7π/6) + 2πn

Θ = (11π/6) + 2πn

(π/6) + π = (7π/6)

(5π/6) + π = (11π/6)

Θ = (π/6) + πn

Θ = (5π/6) + πn

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I hope this helps!

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