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Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Solve for x: tan (x) + sec (x) = 2cos (x)?

Solve for x: tan (x) + sec (x) = 2cos (x)

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  • Puggy
    Lv 7
    1 decade ago
    Favorite Answer

    tan(x) + sec(x) = 2cos(x)

    Change everything to sines and cosines.

    [sin(x)/cos(x)] + [1/cos(x)] = 2cos(x)

    Get rid of all fractions by multiplying both sides by cos(x).

    sin(x) + 1 = 2cos^2(x)

    Change cos^2(x) into 1 - sin^2(x) as per the identity.

    sin(x) + 1 = 2[1 - sin^2(x)]

    sin(x) + 1 = 2 - 2sin^2(x)

    Move everything to the left hand side.

    2sin^2(x) + sin(x) + 1 - 2 = 0

    2sin^2(x) + sin(x) - 1 = 0

    Factor the same way you would factor 2y^2 + y - 1.

    In fact, let me make it clear for you by using the substitution y = sin(x).

    2y^2 + y - 1 = 0

    (2y - 1)(y + 1) = 0

    Which means

    2y - 1 = 0

    2y = 1

    y = 1/2, but back-substitute y = sin(x), to get

    sin(x) = 1/2, so

    x = { pi/6, 5pi/6 }

    y + 1 = 0

    y = -1

    sin(x) = -1

    x = 3pi/2

    Therefore,

    x = {pi/6, 5pi/6, 3pi/2 }

    However, we should determine whether all of these solutions work, by plugging them into the original.

    tan(x) + sec(x) = 2cos(x)

    Test x = pi/6:

    tan(pi/6) + sec(pi/6) = 2cos(pi/6)?

    [1/sqrt(3)] + [2/sqrt(3)] = 2sqrt(3)/2 ?

    [3/sqrt(3)] = sqrt(3) ?

    3sqrt(3)/3 = sqrt(3) ?

    sqrt(3) = sqrt(3) ? YES.

    Keep this solution.

    Test x = 5pi/6.

    tan(5pi/6) + sec(5pi/6) = 2cos(5pi/6) ?

    [-1/sqrt(3)] + [-2/sqrt(3)] = 2[-sqrt(3)/2] ?

    -3/sqrt(3) = -sqrt(3) ?

    -3sqrt(3)/3 = -sqrt(3) ?

    -sqrt(3) = -sqrt(3) ? YES.

    Test x = 3pi/2:

    tan(x) + sec(x) = 2cos(x)

    tan(3pi/2) + sec(3pi/2) = 2cos(3pi/2)

    This one fails, because tan(3pi/2) is undefined.

    Reject x = 3pi/2.

    Making the only solutions

    x = {pi/6, 5pi/6}

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  • Philo
    Lv 7
    1 decade ago

    tan x + sec x = 2cos x

    sin/cos + 1/cos = 2cos

    sin + 1 = 2 cos²

    sin + 1 = 2( 1 - sin²)

    sin + 1 = 2 - 2sin²

    2sin² + sin - 1 = 0

    (2sin - 1)(sin + 1) = 0

    sin x = 1/2, x = π/6 or x = 5π/6

    sin x = -1, x = 3π/2

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  • 5 years ago

    This Site Might Help You.

    RE:

    Solve for x: tan (x) + sec (x) = 2cos (x)?

    Solve for x: tan (x) + sec (x) = 2cos (x)

    Source(s): solve tan sec 2cos x: https://tr.im/Ht6P3
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  • Anonymous
    4 years ago

    For the best answers, search on this site https://shorturl.im/avbue

    Pretty sure you can't divide by your du. But another way to approach it is expand it into 3/sec^2(x)+tan^2(x)/sec^2(x) sub trig id tan^2(x)=sec^2(x)-1 3/sec^2(x)+(sec^2(x)-1)/sec^2(x) expand second, 3/sec^2(x)+(sec^2(x)/sec^2(x))-1/sec^2(... 3/sec^2(x)-1/sec^2(x)+(sec^2(x)/sec^2(x... 2/sec^2(x)+1 sub trig id sec^2=1/cos^2(x) 2cos^2(x)+1 sub trig id cos^2(x)=(1/2)(1+cos2x) 2(1/2)(1+cos2x) +1 1+1+cos2x integrate 2+cos2x=2x+(1/2)sin2x Hope this helps

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  • 4 years ago

    Tanx Secx 2cosx

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  • Chris
    Lv 4
    1 decade ago

    x = 30 degrees

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