trigonometric equation?

ignore whatever is written with pencil. its question 8. If you have time, can you do 8.1 to 8.8 except 8.2 because i have done it.

I know I'm getting too ahead of myself for asking too much to a stranger but i really need to know to do this before exam.

pls ONLY USE COS, SIN and TAN.

Update:

these are the answers:

8.1) x=11pi/30 and pi/30

8.3) pi/2 and 3pi/4

i have already got the answers, i need to know how to do it.

Attachment image

2 Answers

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  • 10 months ago

    8.1: cos(x - pi/5) = sqrt(3)/2 =>

    x - pi/5 = pi/6 or -pi/6

    x = pi/5 + pi/6 or pi/5 - pi/6

    = 11pi/30 or 1pi/30.

    8.3: 2*sin(x)*cos(x) + 2*cos^2(x) = 0 =>

    either cos(x) = 0 or sin(x) = -cos(x) =>

    x = 0 or x = 3pi/4 or x = 7pi/4.

    8.5: 3[2*cos^2(x) - 1] + 5*cos(x) - 1 = 0 =>

    6*cos^2(x) + 5*cos(x) - 4 = 0 =>

    [3*cos(x) + 4]*[2*cos(x) - 1] = 0 =>

    cos(x) = 1/2, because it can't be -4/3.

    So x = pi/3 or 5pi/3.

    8.7: sin(x)cos(pi/6) + cos(x)*sin(pi/6) = 2*cos(x)cos(pi/3) + 2*sin(x)*sin(pi/3)

    =>

    sin(x)*sqrt(3)/2 + (1/2)*cos(x) = 1*cos(x) + sqrt(3)*sin(x) =>

    -(1/2)*cos(x) = sin(x)*sqrt(3)/2 =>

    sin(x)/cos(x) = -1/sqrt(3) =>

    tan(x) = -1/sqrt(3) =>

    x = 5pi/6 or 11pi/6.

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  • ted s
    Lv 7
    10 months ago

    it appears YOU need to LEARN some trig identities...cos ( 2x) = 2 cos² x - 1 = 1 - 2 sin² x = cos² x - sin² x ; cos ( A + B ) = cos A cos B - sin A sin B { if A - B then: ... cos B + sin A ... } ; sin (A ± B ) = sin A cos B ± cos A sin B ; sin (2x) = 2 sin x cos x ;..use these to work some of the problems....such as (8.2) : sin x ( 1 / √2) + cos x ( 1 / √2) = √2 cos x ====> sin x + cos x = 2 cos x ===> sin x = cos x ===> x = π / 4 + n π

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