Find all solutions (if they exist)for the given equation on the interval [0,2pi)
cos^2x-5sinx+5=0
cos^2x-5sinx+5=0
1-sin^2x - 5sinx + 5 = 0
sin^2x + 5sinx - 6 = 0
(sinx+6)(sinx-1) = 0
so,
sinx = -1/6 or 1
Now you can find x. Watch for signs in quadrants.
To find all the solutions for the given equation on the interval [0, 2pi), we can use the following steps:
Step 1: Rewrite the equation in terms of a single trigonometric function. You can use the identity cos^2(x) = 1 - sin^2(x) to replace cos^2(x) in the equation:
1 - sin^2(x) - 5sin(x) + 5 = 0
Step 2: Rearrange the equation to a quadratic form. Move all terms to one side to obtain:
-sin^2(x) - 5sin(x) + 6 = 0
Step 3: Factor the quadratic equation. Since the equation is in terms of sin(x), we can factor it:
(-sin(x) - 2)(sin(x) + 3) = 0
Step 4: Set each factor equal to zero and solve for x:
-sin(x) - 2 = 0 or sin(x) + 3 = 0
Step 5: Solve for x in each equation separately:
For -sin(x) - 2 = 0:
sin(x) = -2
There are no solutions for sin(x) = -2 on the interval [0, 2pi) since the sine function ranges from -1 to 1.
For sin(x) + 3 = 0:
sin(x) = -3
Again, there are no solutions for sin(x) = -3 on the interval [0, 2pi).
Therefore, the given equation has no solutions on the interval [0, 2pi).