Please solve the eqution Cos4x + cos2x - 2 =0

Mathematics

1 answer:

horrorfan [7]3 years ago

6 0

cos4x = cos2x

We know that:

cos2x = 1-2cos^2 x

==> cos4x = 1-2cos^2 (2x)

Now substitute:

==> 1-2cos^2 (2x) = cos2x

==> 2cos^2 (2x) + cos2x - 1 = 0

Now factor:

==> (2cos2x -1)(cos2x + 1) = 0

==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3

==> x1= pi/6 , 7pi/6

==> x1= pi/6 + 2npi

==> x2= 7pi/6 + 2npi

==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.

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The equation line l is given by the equation y= 7/4x+3. line m goes through the point ( –1,4) and is perpendicular to line l whi

mr_godi [17]

We determine line m as follows:

*First, by theorem we have the following:

$m_1=-\frac{1}{m_2}$

Here m1 & m2 are the slopes of two perpendicular lines. For all lines that are perpendicular that is true, so we calculate the slope of line m using the slope of the function given [Which has a slope of 7/4]:

$m_1=-\frac{1}{\frac{7}{4}}\Rightarrow m_1=-\frac{4}{7}$

So, the slope of line m is -4/7. Now, using this slope and the point (-1, 4) we replace in the following expression:

$y-y_1=m_1(x-x_1)$

Here x1, y1 & m1 are the x-component of the point, the y-component of the point, and the slope of the line respectively, so we replace and solve for y: