solve the following equations for all values of x:
d) 2sin^2(x) = 2 + cosx
f) cos(-2x) + 2 = 3cos(2x)
and if it isnt too much to ask T_T
simplify the following expressions:
b) cos^2x + tan^2xcos^2x
thankss
d) $\displaystyle 2sin^2(x) = 2 + cosx$
Expand the left hand side: $\displaystyle 2(1-cos^2(x))= 2 + cos x$
Move all terms to one side and factorise: $\displaystyle cos(x) (2cos(x)+1) = 0$
I assume you can take it from here?
f) $\displaystyle cos(-2x) + 2 = 3cos(2x)$
Note: cos(-2x) = cos(2x) [proof by unit circle; or by proving cosine graph is an even function]
$\displaystyle cos(2x) + 2 = 3cos(2x)$
$\displaystyle 2(cos(2x)+1) = 0$
Now solve that.
b) $\displaystyle cos^2x + tan^2xcos^2x$
Factorise first $\displaystyle cos^2(x) (1+tan^2(x) )$
Use the identity $\displaystyle tan^2(x) + 1 = sec^2(x)$
$\displaystyle = cos^2(x)sec^2(x)$
= 1
d) Knowing that sin^2 x + cos^2 x = 1, we can say that sin^2 x = 1 - cos^2 x. Now we can replace the 2sin^2 x with 2 - 2cos^2 x:
2 - 2cos^2 x = 2 + cos x
Subtract 2 from both sides:
-2cos^2 x = cos x
2cos^2 x + cos x = 0
(cos^2 x)(2cos^2 x + 1) = 0
cos^2 x = 0 or cos^2 x = -0.5
The second case is not possible, so cos^2 x = 0, which means that:
x = pi * k (or 180 degrees * k).
e) Cosine is an even function, meaning that cos (-x) = cos x. That means cos (-2x) = cos (2x), so:
cos (2x) + 2 = 3cos (2x)
2cos (2x) = 2
cos (2x) = 1
2x = pi / 2 + 2pi * k (or 90 degrees + 360 degrees * k)
x = pi / 4 + pi * k (or 45 degrees + 180 degrees * k).
b) cos^2x + tan^2xcos^2x
tan^2 x = (sin^2 x)/(cos^2 x). Multiply that by cos^2 x, and you'll get sin^2 x.
As said at the beginning of the post, sin^2 x + cos^2 x = 1, so this whole expression equals 1.
nadavs