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C2 Chapter 10 - Trigonometric Identities Question

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    Solve 2sinxcosx+cos^2x= 0

    for 0 < x < 360

    What are the solutions?
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    Move the cos^2x over to the other side. Divide by cosx. Then you should know what to do.
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    (Original post by TanishaBellum)
    Solve 2sinxcosx+cos^2x= 0

    for 0 < x < 360

    What are the solutions?
     2sinxcosx + cos^2x = 0 \implies cosx(2sinx+cosx)=0

    Now we get two equation,  cosx=0 \text{ and } 2sinx+cosx=0 , solve both the equations to find the solutions.
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    i've solved cosx = 0 as 90 and 270 degrees?

    i cant solve the other one :?
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    (Original post by TanishaBellum)
    i've solved cosx = 0 as 90 and 270 degrees?

    i cant solve the other one :?
    You have got two correct solutions.

    For the other one,  2sinx+cosx=0 \implies 2sinx=-cosx

    Now divide both sides by  cosx to get an equation involving only  tanx .
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    (Original post by Math12345)
    Move the cos^2x over to the other side. Divide by cosx. Then you should know what to do.
    You will lose some solutions if you do it in this way. You also need to solve it for  cosx=0
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    (Original post by raheem94)
    You will lose some solutions if you do it in this way. You also need to solve it for  cosx=0
    Thanks for the tip.
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    (Original post by bariskesin)
    Solve 2sinxcosx + cos^2x

    for 0 < x < 360

    What are the solutions?


    I'd start off by using sinx/cosx= tanx
    move cos^2x to the other side.
    you get 2sinxcosx=-cos^2x
    divide through by cos x ---> 2sinx= -1
    sinx= -1/2
    solution one---> shiftsin(-1/2)= -30 degrees
    solution two ---> sine is positive in Q2: 180 degrees- (-30)= 210 degrees
    other solution(s)---> sine graph has period of 360 degrees (.ie. it repeats every 360 degrees) so +/- 360 to both solutions to find other solutions satisfying range. we can only add 360 degrees to the -30 and not to the 210 degrees. Nor can we subtract 360 degrees from either solutions as the values we would obtain would be out of range. Therefore, the solutions within the range of 0<x<360 are 210 and 330 degrees.

    Hopefully I've got this correct
    Unfortunately, you are wrong.
     2sinxcosx=-cos^2x

    Dividing through by  cosx gives,  2sinx = -cosx
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    (Original post by raheem94)
    You have got two correct solutions.

    For the other one,  2sinx+cosx=0 \implies 2sinx=-cosx

    Now divide both sides by  cosx to get an equation involving only  tanx .
    okay..

    im at tanx + sinx + cosx = 0

    do you square everything and then end up with...

    tan^2x + sin^2x + cos^2x = 0



tan^2x + 1 = 0



tan^2x = -1

    root and inverse tan

    x = inverse tan (1)

    quadrant rule and x = 45 + 225 along with 90 + 270

    is this right?
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    (Original post by TanishaBellum)
    okay..

    im at tanx + sinx + cosx = 0

    do you square everything and then end up with...

    tan^2x + sin^2x + cos^2x = 0



tan^2x + 1 = 0



tan^2x = -1

    root and inverse tan

    x = inverse tan (1)

    quadrant rule and x = 45 + 225 along with 90 + 270

    is this right?
     2sinx=-cosx

    Dividing everything by cosx gives,  2tanx=-1 \implies tanx =- \frac12

    Now just find the tan inverse of -0.5 and use the quadrant diagram to find the other solutions.
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    (Original post by raheem94)
     2sinxcosx + cos^2x = 0 \implies cosx(2sinx+cosx)=0

    Now we get two equation,  cosx=0 \text{ and } 2sinx+cosx=0 , solve both the equations to find the solutions.
    how did you get the bracket bit
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    (Original post by raheem94)
     2sinxcosx + cos^2x = 0 \implies cosx(2sinx+cosx)=0

    Now we get two equation,  cosx=0 \text{ and } 2sinx+cosx=0 , solve both the equations to find the solutions.
    is it because cosx is common in both?
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    (Original post by dongonaeatu)
    how did you get the bracket bit
    (Original post by dongonaeatu)
    is it because cosx is common in both?
    I just pulled cosx as a common factor.
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    (Original post by dongonaeatu)
    is it because cosx is common in both?
    Yes. It's always best to try get something = 0 into a product of factors = 0 if possible as then you can tackle each bracket independently being = 0 (like when solving a quadratic)
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    (Original post by raheem94)
    I just pulled cosx as a common factor.

    (Original post by hassi94)
    Yes. It's always best to try get something = 0 into a product of factors = 0 if possible as then you can tackle each bracket independently being = 0 (like when solving a quadratic)
    Thanks

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