keep in mind answers should be pi radians or if necessary 3 dec.
csc^2-2=0
all angles
4cos^2x-4cosx+1=0
all angles and name 8 angles
7sin^2x-22sinx+3=0
all angles and name 8 angles
2cos^2x-7cosx=-3
all angles
sinx(2sinx+1)=0
all angles
1. csc^2 x = 2
sin^2 x = 1/2
sin x = ±1/√2
x = π/4, 3π/4 , 5π/4 , 7π/4
2. 4cos^2x - 4cosx + 1 = 0
(2cos x - 1)^2 = 0
(2cosx - 1)= 0
cosx = 1/2
x = π/3 or x = 5π/3
3.
7sin^2 x - 22sinx + 3 = 0
(7sinx - 1)(sinx - 3) = 0
sinx = 1/7 or sinx = 3 , which is not possible
for sinx = 1/7
make sure calculator is set to radians
x = .1433 (in I) or x = π - .1433 = 2.998 ( in II)
do the rest the same way
yup
To find the solutions to each of these equations, we will apply various methods based on trigonometric identities and algebraic techniques. Let's start with the first equation:
1) csc^2(x) - 2 = 0
To obtain the solutions, we can rewrite the equation in terms of sine:
sin^2(x) - 2cos^2(x) = 0
Since sin^2(x) = 1 - cos^2(x) from the Pythagorean identity, we can substitute:
1 - cos^2(x) - 2cos^2(x) = 0
Now, combine like terms:
-3cos^2(x) + 1 = 0
Solving this quadratic equation for cos(x), we get:
cos(x) = ±√(1/3)
Now, to find the angles, we take the inverse cosine of both sides:
x = cos^(-1)(±√(1/3))
The values between 0 and 2π that satisfy this equation give the angles for the first equation.
2) 4cos^2(x) - 4cos(x) + 1 = 0
We can solve this quadratic equation using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
In this case, a = 4, b = -4, and c = 1. Plugging these values into the quadratic formula and simplifying, we get:
x = [2 ± √(4 - 16)] / 8
x = [2 ± √(-12)] / 8
Since we cannot take the square root of a negative number when dealing with real solutions, there are no real solutions for this equation. However, if we consider complex solutions, we can simplify further:
x = [2 ± √(12)i] / 8
x = [1 ± √(3)i] / 4
Therefore, the equation has purely complex solutions, and those are the angles.
3) 7sin^2(x) - 22sin(x) + 3 = 0
This equation is also solved using the quadratic formula. In this case, a = 7, b = -22, and c = 3. Plugging these values into the quadratic formula and simplifying, we get:
x = [22 ± √(484 - 84)] / 14
x = [22 ± √(400)] / 14
x = [22 ± 20] / 14
Simplifying further gives us:
x = [1 ± √(10)] / 7
Therefore, the equation has two angles: [1 + √(10)]/7 and [1 - √(10)]/7.
4) 2cos^2(x) - 7cos(x) = -3
To simplify this equation, let's rearrange it as:
2cos^2(x) - 7cos(x) + 3 = 0
We can solve this quadratic equation in the same way as before, by applying the quadratic formula. In this case, a = 2, b = -7, and c = 3.
x = [7 ± √(49 - 24)] / 4
x = [7 ± √(25)] / 4
x = [7 ± 5] / 4
Hence, the equation has two solutions: x = 3/2 and x = 3/4.
5) sin(x)(2sin(x) + 1) = 0
To solve this equation, we have two cases to consider:
Case 1: sin(x) = 0
If sin(x) = 0, then x must be equal to 0 radians or π radians.
Case 2: 2sin(x) + 1 = 0
Solving this linear equation for sin(x), we get:
2sin(x) = -1
sin(x) = -1/2
This implies that x must be equal to 7π/6 or 11π/6.
In summary, the solutions for each equation are as follows:
1) csc^2(x) - 2 = 0: angles = cos^(-1)(±√(1/3))
2) 4cos^2(x) - 4cos(x) + 1 = 0: no real solutions, only complex solutions
3) 7sin^2(x) - 22sin(x) + 3 = 0: angles = [1 ± √(10)]/7
4) 2cos^2(x) - 7cos(x) = -3: angles = 3/2 and 3/4
5) sin(x)(2sin(x) + 1) = 0: angles = 0, π, 7π/6, and 11π/6.