The solutions to the equation on the given interval are;
x = π/4, 3π/4, 5π/4, 7π/4.
What is the solution to the equation on the interval?
Given that;
4(sin x)² - 2 = 0
Add 2 to both sides and divide both sides 4
4(sin x)² - 2 + 2 = 0 + 2
4(sin x)² = 2
(4(sin x)²)/4 = 2/4
(sin x)² = 1/2
Square both sides
√((sin x)²) = ±√(1/2)
sin x = ±√(1/2)
sin x = ±(√2)/2
Next, we solve for x
Not that 180° = π
x = sin⁻¹ ( (√2)/2 ) = 45° = 180°/4 = π/4
x = π/4
Since the sine function is positive in the first and second quadrant, we subtract the reference angle from π to find the solution in the second quadrant.
x = π - π/4
x = 3π/4
Now, we find the period of sin x
2π / |b|
We know that, the distance between a number and zero is 1
2π / 1
2π
Hence, period of sin x function is 2π, values will repeat every 2π radians in both direction.
Since sine function is negative in third and fourth quadrant,
x = 2π + π/4 + π
x = 5π/4
Now, add 2π to every negative angle to get a positive angle
x = 2π + ( - π/4 )
x = 2π×4/4 - π/4
x = ((2π × 8) - π ))/4
x = (8π - π)/4
x = 7π/4
Therefore, the solutions to the equation on the given interval are;
x = π/4, 3π/4, 5π/4, 7π/4.
Learn more on radian solutions here: https://brainly.com/question/16044749
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