1. ## solved algebracially

arcsinx-arccosx = 0

i know x is 45

but how to do algebraccly

2. Check the problem.
If sinx - cosx = 0 then x = 45 degrees.
Now
arcsinx+arccosx = π/2

arcsinx-arccosx = arcsinx+arccosx - 2arcosx = 0

π/2 -2arccosx = 0

Solve for x.

3. Hello, aeroflix!

Solve: .$\displaystyle \arcsin x-\arccos x \:=\: 0$

i know x is 45 . .
This is wrong!

$\displaystyle \text}Let: }\,\alpha = \arcsin x \quad\Rightarrow\quad \begin{Bmatrix}\sin\alpha \:=\: x \\ \\[-3mm] \cos\alpha \:=\: \sqrt{1-x^2} \end{Bmatrix}$ .[1]

$\displaystyle \text{Let: }\,\beta = \arccos x \quad\Rightarrow\quad \begin{Bmatrix} \cos\beta \:=\:x \\ \\[-3mm] \sin\beta \:=\:\sqrt{1-x^2} \end{Bmatrix}$ .[2]

The equation becomes: .$\displaystyle \alpha - \beta \;=\;0$

Take the sine of both sides: .$\displaystyle \sin(\alpha - \beta) \;=\;\sin 0$

and we have: .$\displaystyle \sin\alpha\cos\beta - \cos\alpha\sin\beta \;=\;0$

Substitute [1] and [2]: .$\displaystyle (x)(x) - (\sqrt{1-x^2})(\sqrt{1-x^2}) \;=\;0$

. . $\displaystyle x^2 - (1-x^2) \:=\:0 \quad\Rightarrow\quad 2x^2-1 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:\frac{1}{2}$

Therefore: .$\displaystyle x \;=\;\pm\dfrac{1}{\sqrt{2}}$

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I tried to check my answers and ran into a new puzzle.
. . (New to me, anyway.)

First, I assumed that the angles are in $\displaystyle [0^o,\:360^o]$

Then I tested $\displaystyle x = \frac{1}{\sqrt{2}}$

We have: .$\displaystyle \arcsin\left(\frac{1}{\sqrt{2}}\right) - \arccos\left(\frac{1}{\sqrt{2}}\right)\quad\hdots$ .Does it equal 0 ?

We know that: .$\displaystyle \begin{Bmatrix}\arcsin(\frac{1}{\sqrt{2}}) &=& 45^o,\:135^o \\ \\[-3mm] \arccos(\frac{1}{\sqrt{2}}) &=& 45^o,\:315^o \end{Bmatrix}$

So we have: .$\displaystyle \begin{Bmatrix}45^o \\ 135^o\end{Bmatrix} - \begin{Bmatrix}45^o \\ 315^o\end{Bmatrix} \;\begin{array}{c}_{_?} \\=\\ ^{^?}\end{array} \; 0$

The equation is true if we select $\displaystyle 45^o$ from both sets.

Query: .Are we allowed to do that?