MATHEMATICS MIDDLE SCHOOL

2 sin^2 (x) -5 sin (x) -3=0 I. Rewrite the equation by substituting the expression u in for sin x.

II. Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.

III. Use the zero product property to solve the quadratic equation.

IV. Rewrite your solutions to Part III by replacing u with sin x.

V. Solve the remaining equations for x, giving all solutions to the equation.

Answers

Answer 1

Answer:

we have that

$2sin^(2) x-5sin x-3=0$

I. Rewrite the equation by substituting the expression u in for sin x.

$2u^(2) -5u-3=0$

II. Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.

$2u^(2) -5u-3=0$ is equal to

using a graph calculator-----> see the attached figure

$(u-3)*(2u+1)=0$

III. Use the zero product property to solve the quadratic equation.

$(u-3)*(2u+1)=0$

(u-3)=0--------------> u=3

(2u+1)=0-------- 2u=-1------> u=-1/2-----> u=-0.5

IV. Rewrite your solutions to Part III by replacing u with sin x.

sin x=3--------> is not the solution (sin x can not be greater than 1)

sin x=-0.50------>is the solution

V. Solve the remaining equations for x, giving all solutions to the equation.

sin x=-0.50

if the sine is negative

then

x belong to the III or IV quadrant

we know that

sin 30°=0.50

the solution for the III quadrant is

x=180°+30°-------> x=210°

the solution for the IV quadrant is

x=360°-30°------> x=330°

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Answers

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Answers

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Answers

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Answers

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Step-by-step explanation:

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Answers

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