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`sinx/sqrt 2 = cos^2x`

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schlule | Student, Grade 11 | eNotes Newbie

Posted May 16, 2013 at 12:13 AM via web

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`sinx/sqrt 2 = cos^2x`

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 16, 2013 at 4:39 AM (Answer #4)

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`u_1=-1` and` u2=0.5`


`0.5=cos^2x_2-gtx_2= pi/4`

The second root still does not exist for cosine, so we can convert and end up with the same result as oldnick




You will find that if you try to calculate `arcsinsqrt2`  in your calculator you will get an error.  That is because it does not exist.  The values that exist for arcsin fall between -1 and 1, and `sqrt(2)>1`

The value of `sin((3pi)/4)` is `sqrt(2)/2` ; therefore, that cannot be the solution to `arcsinsqrt2`  .  Rather, it is a multplie of the one existing root `pi/4` .

Also, observe in the graph of the function that the graph does not cross the x-axis at sqrt(2), further supporting that the root does not exist. Rather, all roots of this function are multiples of the 1 existing root, pi/4, as it is an infinitely repeating function.


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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 16, 2013 at 12:36 AM (Answer #1)

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Presumably you are wanting to solve for x?


We know that:


Therefore if we square both sides of equation we wish to solve:


Now substitute the identity above:


Rearranging we have the equation:


In order to solve for the roots of this equation we will simplify by setting `u=cos^2x`



Two numbers that muliply to -0.5 and add to 0.5 are 1 and -0.5.  Therefore:

u1=1 and u2=-0.5

Now we can solve for x1 and x2:

`1=cos^2x_1 -> x_1=0`


As you cannot take the square root of a negative number, x2 does not exist.


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oldnick | (Level 1) Valedictorian

Posted May 16, 2013 at 12:48 AM (Answer #3)

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Root exists, for `x=pi/4 +kpi`  and `x=3/4 +kpi`

Indeed  `sinx=sqrt(2)/2`

and `sinx/sqrt(2)= sqrt(2)/2sqrt(2)=1/2=cos^2x=(sqrt(2)/2)^2)=1/2` !!!!!!!!!!!!!


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