Question

Cos^2x – sin^2x = sinx – cosx

Answer 1

$\sf cos^2x-sin^2x=sinx-cosx \\ (cosx-sinx)(cosx+sinx)+(cosx-sinx)=0 \\ (cosx-sinx)(cosx+sinx+1)=0 \\ \\ cosx-sinx=0 \\ sinx=cosx \\ tgx=1 \\ \boxed{\sf x=\dfrac{\pi}{4}+\pi k } \\ \\ \\ cosx+sinx+1=0$

Применим универсальную тригонометрическую подстановку

$\sf \dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1=0 \\ 1-t^2+2t+1+t^2=0 \\ t=-1 \\ \\ tg\left(\dfrac{x}{2} \right)=-1 \\ \dfrac{x}{2}=-\dfrac{\pi}{4}+\pi k\\ \boxed{\sf x=-\dfrac{\pi}{2}+2\pi k}$

При этом cos(π)+sin(π)+1=-1+0+1=0, значит нужно добавить решение

$\boxed{\sf x=\pi+2 \pi k}$

Ответ:  $\left [ \begin{array}{I} \sf x=\dfrac{\pi}{4}+\pi k \\ \sf x=\pi+2 \pi k\\ \sf x=-\dfrac{\pi}{2}+2\pi k \end{array}; \sf \ k \in \mathbb{Z}$

Answer 2

                Решение : ////////////////////////////////////

Answer 3

We can use the trigonometric identity:

cos^2x - sin^2x = cos(2x)

So the equation becomes:

cos(2x) = sinx - cosx

We can rewrite sinx - cosx as:

sqrt(2)*sin(pi/4 - x)

So the equation becomes:

cos(2x) = sqrt(2)*sin(pi/4 - x)

Using the double angle identity for cosine, we get:

2cos^2x - 1 = sqrt(2)*sin(pi/4 - x)

Using the identity sin(a - b) = sin(a)cos(b) - cos(a)sin(b), we can rewrite the right-hand side as:

sqrt(2)/2(cos(x) + sin(x))

So the equation becomes:

2cos^2x - 1 = sqrt(2)/2(cos(x) + sin(x))

Multiplying both sides by 2, we get:

4cos^2x - 2 = sqrt(2)(cos(x) + sin(x))

Multiplying both sides by sqrt(2) and simplifying, we get:

2sqrt(2)cos^2x - sqrt(2)cosx - sqrt(2)sinx - 2sqrt(2) = 0

Dividing both sides by sqrt(2) and simplifying, we get:

2cos^2x - cosx - sinx - 2 = 0

Using the identity cosx = sqrt(1 - sin^2x), we can rewrite the equation as a quadratic equation in sinx:

2(1 - sin^2x) - sqrt(1 - sin^2x) - sinx - 2 = 0

Letting y = sinx, we can rewrite the equation as:

2(1 - y^2) - sqrt(1 - y^2) - y - 2 = 0

This equation can be solved using numerical methods or approximated using a graphing calculator or software.