Consider equation #5/2cos(2x)-1/2=3sin(x)# a) Put the equation into standard quadratic trigonometric equation form. b) Use the quadratic equation to factor the equation. c) What are the solutions to the equation to decimal places, where #0<=x<=360^@#?

2 Answers
Douglas K.
Dec 17, 2017

a) Put the equation into standard quadratic trigonometric equation form.

#5/2cos(2x)-1/2=3sin(x)#

Multiply both sides of the equation by 2:

#5cos(2x)-1=6sin(x)#

Substitute #cos(2x) = 1-2sin^2(x)#

#5(1-2sin^2(x))-1=6sin(x)#

Use the distributive property:

#5-10sin^2(x)-1=6sin(x)#

Combine like terms and move the sine function to left:

#-10sin^2(x)-6sin(x)+4 =0#

Divide both sides by -2:

#5sin^2(x)+3sin(x)-2 =0 color(red)(larr"this is standard form.")#

b) Use the quadratic equation to factor the equation.

The quadratic equation adapted for the form, #asin^2(x)+bsin(x)+c=0#, is:

#sin(x) = (-b+-sqrt(b^2-4(a)(c)))/(2a)#

where #a = 5, b = 3, c = -2#

#sin(x) = (-3+-sqrt(3^2-4(5)(-2)))/(2(5))#

#sin(x) = (-3+-sqrt(3^2-4(5)(-2)))/(2(5))#

#sin(x) = 2/5# and #sin(x)= -1#

I will leave part c) for you to do using your calculator.

Ratnaker Mehta
Dec 17, 2017

#"The Solution Set "sub [0^@,360^@]="{23.58^@,156.42^@,270^@}.#

Explanation:

#5/2*cos2x-1/2=3sinx, or, 5cos2x-1=6sinx#.

Since, #cos2x=1-2sin^2x#, we have,

#5(1-2sin^2x)-1=6sinx, i.e., #

#10sin^2x+6sinx-4=0, or,#

#5sin^2x+3sinx-2=0#.

#sinx=trArr 5t^2+3t-2=0#.

Using Quadratic Formula, we get,

#t=sinx={-3+-sqrt(3^2-4*5*(-2))}/(2*5)#.

#:. sinx=(-3+-7)/10#.

#:. sinx=0.4, or, sinx=-1.#

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

#:. sinx=0.4=sin23.58^@, or, x=180^@-23.58^@=156.42^@#.

#sinx=-1=sin270^@#.

#:."The Solution Set "sub [0^@,360^@]="{23.58^@,156.42^@,270^@}.#

Enjoy Maths.!

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