Solve for x in the interval [-pi,0]
a) sin^2x = 3/4
I know that you have +root3/2 and
-root3/2 and the positive one gives you an error when doing the inverse of sin, but im confused about the -root3/2.
I found that one of the answers of x is
-pi/3 (-60 degrees) which fits my interval, but I don't know how you get
-2pi/3 as the other answer.
I did -pi/3 + pi but -2pi/3 as a positive angle which does not fit into the interval.
Please explain how you get the answer -2pi/3?
as you say, sinx = ±√3/2
You know the reference angle is π/3.
sinx > 0 in QI, QII, so you have π/3 and 2π/3
sinx < 0 in QIII, QIV, so that gives you 4π/3 and 5π/3
But, for this problem, the domain is [-π,0] (QIII and QIV), so you have to subtract 2π giving you -π/3 instead of 5π/3, and -2π/3 instead of 4π/3.
Trigonometric identity:
sin ( - x ) = - sin x
On interval [ 0 , pi ] [ 0 , 180 ° ]
you have 2 angles with sin x = sqroot ( 3 ) / 2
x = pi / 3 = 60 °
and
x = 2 pi / 3 = 120 °
So if sin ( - x ) = sin x
on interval [ - pi , 0 ] [ - 180 ° , 0 ]
you have 2 angles with sin x = sqroot ( 3 ) / 2
x = - pi / 3 = - 60 °
and
x = - 2 pi / 3 = - 120 °
To find the value of x in the interval [-π, 0] that satisfies the equation sin^2x = 3/4, let's take a step-by-step approach:
1. First, take the square root of both sides of the equation to eliminate the square:
√(sin^2x) = √(3/4)
2. This simplifies to:
|sin x| = √(3/4)
3. Since sin x can either be positive or negative, we need to consider both cases.
Case 1: sin x is positive.
In the interval [-π, 0], sin x is positive in the second quadrant and the fourth quadrant.
In the second quadrant, sine is positive and the reference angle is π - x. Therefore, we have:
sin (π - x) = √(3/4)
Solving for x:
π - x = arcsin(√(3/4))
π - x = π/3
x = -2π/3
Case 2: sin x is negative.
In the interval [-π, 0], sin x is negative in the third quadrant.
In the third quadrant, sine is negative, and the reference angle is x - π. Therefore, we have:
sin (x - π) = √(3/4)
Solving for x:
x - π = arcsin(√(3/4))
x - π = π/3
x = π + π/3
x = 4π/3
4. Now, we have two possible values for x, which are -2π/3 and 4π/3. However, we need to ensure that they lie within the specified interval [-π, 0].
-2π/3
Since -2π/3 is less than -π, it lies within the interval.
4π/3
To express 4π/3 within the interval [-π, 0], you can subtract 2π from it:
4π/3 - 2π = -2π/3
Therefore, 4π/3 is equivalent to -2π/3 and also lies within the interval.
Hence, the solutions for x in the interval [-π, 0] satisfying sin^2x = 3/4 are -2π/3 and -2π/3.