Answer:
please see detailed answer below
Explanation:
Trig. identity sin²x + cos²x = 1, cos²x = 1 - sin²x
3 sin²x + (1 - sin²x) - 5 = 7 sinx
subtract 7 sin x from both sides:
3 sin²x + 1 - sin²x - 5 - 7 sinx = 0
2 sin²x - 7 sin x - 4 =0
we have a quadratic.
use the quadratic formula but using sin x instead of x:
x = ((-b ± √(b² - 4ac)) ÷ 2a)
where a is the value of the first coefficient, b is value of the second and c is value of the constant.
sin x = [(-7 ± √((-7)² - 4(2)(-4))) ÷ 2(2)]
= [(7 ± √(49 + 32)) ÷ 4]
= [7 ± √81] / 4
= (7 ± 9) / 4
= 4 or -0.5.
that is sin x = 4 or sin x = -0.5.
sin x = 4 is invalid.
sin x = -0.5
x = arcsin -0.5
= -30