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

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- May 15th 2008, 06:38 AMphthiriasistrig equations
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 - May 15th 2008, 06:49 AMGusbob
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 - May 15th 2008, 06:51 AMnadavs
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