How do you solve $2 cos^2(4x)-1=0$ ?

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How do you solve $2 cos^2(4x)-1=0$ ?

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Answer 1 · 2016-07-25T15:12:36

The Soln. Set $={kpi/2+-pi/16} uu {kpi/2+-3pi/16}, k in ZZ$ .

Explanation:

We recall that the soln. to the eqn. $costheta=cosalpha$ is

$theta=2kpi+-alpha, where, k in ZZ$

Now, the given eqn. is $2cos^2(4x)-1=0 rArr cos4x=+-1/sqrt2$

$:. cos4x=1/sqrt2=cospi/4 rArr 4x=2kpi+-pi/4 rArr x=kpi/2+-pi/16$ , &,

$cos4x=-1/sqrt2=cos3pi/4 rArr 4x=2kpi+-3pi/4$

$rArr x=kpi/2+-3pi/16$

Thus, the Soln. Set $={kpi/2+-pi/16} uu {kpi/2+-3pi/16}, k in ZZ$ .

Answer 2 · 2016-07-25T17:37:39

$pi/16 + (kpi)/4$
$(3pi)/16 + (kpi)/4$

Explanation:

Apply the trig identity:
$cos 2x = 2cos^2 x - 1$
$cos 8x = 2cos^2 (4x) - 1 = 0$
$cos 8x = 0$
Trig table and unit circle give 2 solutions:
a. $8x = pi/2 + 2kpi$
$x = pi/16 + (2kpi)/8 = pi/16 + (kpi)/4$
b. $8x = (3pi)/2 + 2kpi$
$x = (3pi)/16 + (kpi)/4$

Check.
$x = pi/16$ --> $4x = pi/4$ --> $2cos^2 pi/4 = 2(1/2) - 1 = 0$ .OK
$x = (3pi)/16 -> 4x = (3pi)/4 --> 2cos^2 ((3pi)/4) = 2(-sqrt2/2)^2 - 1 = 0$ OK

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How do you solve $2 cos^2(4x)-1=0$ ?

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