Hello. I'm having trouble with the 2 last calculus exercises witch I just can't solve. If someone would please tell me how to solve these two:

1) 1-sinx -cosx +sin2x+cos2x=0

2)log(2 below) (4^(x+1)+4) * log(2below) (4^(x)´+1)=log(1/square root of 2 below) (square root of 1/8)

Thanks to everyone who can help me!!!

2. Does anyone have any ideas or tips how to solve these two, I'm really running out of time

$2)\;\;\log_2\left(4^{x+1}+4\right)\cdot \log_2\left(4^x+1\right) \;=\;\log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1} {8}}\right)$

The right side is: . $\log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1}{8}}\ right) \:=\:p \quad\Rightarrow\quad \left(\frac{1}{\sqrt{2}}\right)^p \:=\:\frac{1}{\sqrt{8}} \quad\Rightarrow\quad p \:=\:3$

The left side is: . $\log_2\bigg[4(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;\bigg[\log_2(4) + \log_2(4^x + 1)\bigg]\cdot\log_2(4^x+1)$

. . $\;=\bigg[2 + \log_2(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;2\cdot\log_2(4^x+1) + \bigg[\log_2(4^x + 1)\bigg]^2$

The equation becomes: . $\bigg[\log_2(4^x+1)\bigg]^2 + 2\!\cdot\!\bigg[\log_2(4^x+1)\bigg]\:=\:3$
. . . . $\bigg[\log_2(4^x+1)\bigg]^2 + 2\bigg[\log_2(4^x+1)\bigg] - 3 \;=\;0$

Factor: . $\bigg[\log_2(4^x+1) +3\bigg]\cdot\bigg[\log_2(4^x+1) -1\bigg] \;=\;0$

$(1)\;\;\log_2(4^x+1) + 3 \:=\:0\quad\Rightarrow\quad \log_2(4^x+1) \:=\:-3 \quad\Rightarrow\quad 4^x + 1 \:=\:2^{-3}$

. . . . $4^x + 1 \:=\:\frac{1}{8} \quad\Rightarrow\quad 4^x \:=\:-\frac{7}{8}$ . . . no real roots

$(2)\;\;\log_2(4^x+1) - 1 \:=\:0 \quad\Rightarrow\quad \log_2(4^x+1) \:=\:1 \quad\Rightarrow\quad 4^x + 1 \:=\:2^1$

. . . . $4^x \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:0}$

I'll assume the answers are on the interval [ $0,\:2\pi)$

$1)\;\;1-\sin x -\cos x +\sin2x+\cos2x\:=\:0$
Double-angle identities: . $\begin{array}{ccc}\sin2A &=& 2\sin A\cos A \\ \cos2A &=& 2\cos^2\!A - 1 \end{array}$

The equation becomes: . $1 - \sin x - \cos x + 2\sin x\cos x + 2\cos^2\!x - 1 \;=\;0$

. . and we have: . $2\sin x\cos x + 2\cos^2\!x - \sin x - \cos x \;=\;0$

Factor: . $2\cos x(\sin x + \cos x) - (\sin x + \cos x) \;=\;0$

Factor: . $(\sin x + \cos x)(2\cos x - 1) \;=\;0$

$(1)\;\;\sin x + \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:-1$

. . . . $\tan x \:=\:-1 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}}$

$(2)\;\;2\cos x - 1 \:=\:0\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2} \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{3},\:\frac{5\pi}{3}}$

5. Soroban, thank you so much, you just made my day!! After being absent from school because of an illness I fell behind and I absolutely could not figure those two out by myself, so A really big thanks to you for helping me out!!