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What are the solutions to sin^2β - sinβ - 2=0 on the interval 0° ≤ β ≤ 360°?

Answer 1

Nolan Alexander

# beta=270^@.#

#sin^2beta-sinbeta-2=0.#

#:. ul(sin^2beta-2sinbeta)+ul(sinbeta-2)=0.#

#:. sinbeta(sinbeta-2)+1(sinbeta-2)=0.#

#:. (sinbeta-2)(sinbeta+1)=0.#

#:. sinbeta=2," which is not possible, as "|sinbeta|le1,# or,

# sinbeta=-1=sin(-pi/2).#

Since, #sintheta=sinalpha rArr theta=(-1)^nalpha+npi, n in ZZ,#

#beta=(-1)^n(-pi/2)+npi, n in ZZ.#

But, we require #beta in [0^@,360^@], beta=3pi/2=270^@.#

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Answer 2

Victoria Adebayo

To find the solutions to the equation sin^2β - sinβ - 2 = 0 on the interval 0° ≤ β ≤ 360°, you can use the quadratic formula. First, let x = sinβ. Then, the equation becomes x^2 - x - 2 = 0. Apply the quadratic formula: x = [ -b ± √(b^2 - 4ac) ] / (2a). Plug in the values: a = 1, b = -1, and c = -2. Solve for x and then find β by taking the arcsin of the solutions. Ensure that the solutions lie within the given interval 0° ≤ β ≤ 360°.

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Answer from HIX Tutor

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.