Two problems stand before you, they say they will beat you to a pulp unless you can solve them. Luckily for you, you are a problem solving kinda dude and aint afraid of no math problems.

Problem one is really mean, he gives you two parts he tells you:

Rewrite the following as one simplified logarithim
a. ln sec x + ln sin x
b. ln (tan^2x+1) + lan cos^2x)

Problem two is the kindest of the two problems he doesn't hold any ill will towards you. He just wants to impress the first problem. He says
Solve the equation for x on the interval [0,2pi) :2sin^2 = 2 + cos x.

Question

Two problems stand before you, they say they will beat you to a pulp unless you can solve them. Luckily for you, you are a problem solving kinda dude and aint afraid of no math problems.

Problem one is really mean, he gives you two parts he tells you: 

Rewrite the following as one simplified logarithim
a. ln    sec x   + ln      sin x    
b. ln (tan^2x+1) + lan cos^2x)

Problem two is the kindest of the two problems he doesn't hold any ill will towards you. He just wants to impress the first problem. He says
Solve the equation for x on the interval [0,2pi) :2sin^2 = 2 + cos x.

cwrw238 · Answer

a .  by laws of logs this is ln (sec x sin x) = ln (sinx/ cos x) = ln tan x

cwrw238 · Answer

b. by laws of logs

= ln (cos^2x (tan^2x + 1)
= ln (cos^2x * sec^2x)
= ln (cos^2x / cos^2 x) 
= ln 1
=  0

cwrw238 · Answer

:2sin^2x  = 2 + cos x.
now sin^2 x = 1 - cos^2 x
so
2(1 - cos^2 x) = 2 + cos x
2 - 2cos^2 x = 2 + cos x
2 cos^2 x + cos x = 0
cos x(2 cos x + 1) = 0
cos x = 0 or -0.5
x = pi/2, 3pi/4, 2pi/3. 4pi/3