Given that sin (x +30)° = cos 2x° for 0°, 0° ≤ x ≤ 90° Find the value x. Hence find the value of cos^2 3x°.
To find the value of x, we can start by using the identity sin (a + b) = sin a * cos b + cos a * sin b.
In this case, we have sin (x + 30)° = cos 2x°, which can be rewritten as sin x * cos 30° + cos x * sin 30° = cos 2x°.
Since cos 30° = √3 / 2 and sin 30° = 1 / 2, we can substitute those values into the equation:
sin x * (√3 / 2) + cos x * (1 / 2) = cos 2x°
Multiplying through by 2 to eliminate the denominators, we get:
√3 * sin x + cos x = 2 * cos 2x°
Using the Pythagorean identity, cos^2 x + sin^2 x = 1, we can rewrite the equation as:
√3 * sin x + cos x = 2 * (1 - sin^2 2x°)
Expanding the right side:
√3 * sin x + cos x = 2 - 2sin^2 2x°
Rearranging the equation, we have:
2sin^2 2x° + √3 * sin x + cos x - 2 = 0
Now, we need to solve this quadratic equation for sin x.
To solve the quadratic equation, we can use the quadratic formula: x = [-b ± √(b^2 - 4ac)] / (2a).
In this case, a = 2, b = √3, and c = 1. Plugging these values into the formula, we have:
sin x = [-√3 ± √(√3^2 - 4 * 2 * 1)] / (2 * 2)
simplifying further:
sin x = [-√3 ± √(3 - 8)] / 4
sin x = [-√3 ± √(-5)] / 4
Since the square root of a negative number is not defined in the real number system, this means that there are no solutions for sin x in this case.
Therefore, there is no specific value of x that satisfies the equation sin (x + 30)° = cos 2x° for 0° ≤ x ≤ 90°.
Without a specific value for x, we cannot directly find the value of cos^2 3x°.