How do you solve #2cos^2x = sinx + 1# from #[0,2pi]#?
x=
(2sinx -1)(sinx +1)=0
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- How do you write #2sin2cos2# as a single trigonometric function?
- How do you write #cos^2 0.45-sin^2 0.45# as a single trigonometric function?
- Prove that # sin^2x + sin^2xcot^2x -= 1# is an identity?
- How do you solve #Cos 2 theta = cos theta#?
- How do you solve for the exact solutions in the interval [0,2pi] of #cosx=sin2x#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7