find the general solutions to the equations:
1) sec x = -2
2) 2 sin^2 x = 1
3) cos^2 x - 2cos x + 1 = 0
a). secx=-2
2pi/3 and 4pi/3
b). sinx=(2^(1/2)/2
its that value at pi/4, and 3pi/4
c). cosx=+ or - 1, so if they want multiple answers its 0, 2pi, and pi
To find the general solutions to the equations, let's break it down step by step:
1) sec x = -2
First, recall that sec x is the reciprocal of cos x. So we can rewrite the equation as 1/cos x = -2. To solve this, we need to find the values of x where the cosine function equals -1/2.
The unit circle shows that cosine is negative in the second and third quadrants. In both quadrants, the cosine value is -1/2 at two points: 2π/3 and 4π/3.
Therefore, the general solutions for x can be written as:
x = 2π/3 + 2πn where n is an integer
x = 4π/3 + 2πn where n is an integer
2) 2 sin^2 x = 1
Divide both sides by 2 to simplify the equation: sin^2 x = 1/2.
To solve this equation, we need to determine the values of x for which the sine function equals the square root of 1/2. The sine function equals ±sqrt(1/2) at two points in each period. We can refer to the unit circle or trigonometric identities to find these values.
Given that sin(x) = sqrt(1/2), we have two solutions within the interval [0, 2π):
x₁ = π/4 and x₂ = 3π/4
Since the sine function has a period of 2π, we can express the general solutions as follows:
x = π/4 + 2πn where n is an integer
x = 3π/4 + 2πn where n is an integer
3) cos^2 x - 2cos x + 1 = 0
This equation is a quadratic in terms of cos x. To solve it, we can factor or use the quadratic formula.
Factor the equation: (cos x - 1)(cos x - 1) = 0
Simplify the expression: (cos x - 1)^2 = 0
Now we have a single factor equal to zero. To find the solutions, set each factor equal to zero and solve for x:
cos x - 1 = 0
Solving for cos x, we have cos x = 1.
Since the cosine function equals 1 at 0 and 2π, we can express the general solutions as:
x = 0 + 2πn where n is an integer
x = 2π + 2πn where n is an integer
So these are the general solutions to the given equations.