Solve cos x-1 = sin^2 x
Find all solutions on the interval [0,2pi)
a. x=pi, x=pi/2, x= 2pi/3
b. x=3pi/7, x=pi/2, x=2pi/3
c. x=3pi/7, x=3pi/2, x=3pi/2
d. x=pi, x=pi/2, x=3pi/2
Using sin^2 x = 1-cos^2(x)
the equation becomes:
cos²(x)+cos(x)-2=0
Use the substitution c=cos(x) to transform the equation to :
c²+c-2=0
(c+2)(c-1)=0
c=-2 or c=+1
Since cos(x) cannot equal -2, solution is rejected.
Now solve for all values of 0≤x≤2π where cos(x)=1
To solve the equation cos(x) - 1 = sin^2(x), we can start by using the trigonometric identity sin^2(x) = 1 - cos^2(x).
Substituting this identity into the equation, we have:
cos(x) - 1 = 1 - cos^2(x)
Rearranging the equation, we get:
cos^2(x) + cos(x) - 2 = 0
Now, let's factorize the quadratic equation:
(cos(x) + 2)(cos(x) - 1) = 0
From here, we can set each factor to zero and solve for x:
1) cos(x) + 2 = 0:
cos(x) = -2
The cosine function ranges from -1 to 1, so there are no solutions for cos(x) = -2 on the interval [0, 2π).
2) cos(x) - 1 = 0:
cos(x) = 1
The cosine function is equal to 1 when the angle is 0 or multiples of 2π. Therefore, the solutions for cos(x) = 1 on the interval [0, 2π) are x = 0 and x = 2π.
In summary, the equation cos(x) - 1 = sin^2(x) has two solutions on the interval [0, 2π): x = 0 and x = 2π.
Therefore, none of the given options (a, b, c, d) correctly represent the solutions to the equation.