Question

1. а) sin(x-1)=cos(x+2) б) sin(3x+3)=cos(x-1) 2. а)8sin² x/2-3sin x-4=0 б)4sin² x/2-cos² x/2=1,5+sin x 3. tg(x-15°)ctg(x+15°)=⅓

Answer 1

$1)$
$a)sin(x-1)=cos(x+2)$
$cos( \frac{ \pi }{2} -x+1)=cos(x+2)$
$\frac{ \pi }{2} -x+1=x+2+2 \pi k$
$-2x=1- \frac{ \pi }{2} +2 \pi k$
$x= \frac{1}{2} ( \frac{ \pi }{2} -1+2 \pi k)$

$b)sin(3x+3)=cos(x-1)$
$sin(3x+3)=sin( \frac{ \pi }{2} -(x-1))$
$3x+3=\frac{ \pi }{2} -(x-1)+2 \pi k и 3x+3= \frac{ \pi }{2} +(x-1)+2 \pi k, kZ.$
$x=\frac{( \frac{ \pi }{2} -2+2 \pi k)}{y}$
$x= \frac{( \frac{ \pi }{2} -4+2 \pi k)}{2} = \frac{ \pi }{4-2+ \pi k}$
$x= \frac{( \frac{ \pi }{2} -2+2 \pi k)}{4}$
$\frac{ x= \pi }{4-2+ \pi k} , kZ$

$2)$
$a)8sin^2 x/2-6sin(x/2)cjs(x/2)-4(sin^2(x/2)+cos^2(x/2))=0$
$4sin^2(x/2)-6sinx/2cosx/2-4cjs^2(x/2)=0$
$4tg^2x/2-6tgx/2-4=0$
$y=tgx/2$      
$4y^2-6y-4=0 2y^2-3y-2=0 y=2y=-0,5 tgx/2=2$
$x/2=arctg2+ \pi k;$
$x=2arctg2+2 \pi k$
$tgx/2=-0,5$
$x=2arctg0,5+ \pi k$

Answer 2

a) To solve sin(x-1)=cos(x+2), we can rewrite it as sin(x-1) - cos(x+2) = 0. Then, using the identity sin(a) - cos(b) = -sqrt(2)*sin(a + pi/4 - b), we have:

-sqrt(2)*sin(x-1 + pi/4 - (x+2)) = 0

which simplifies to:

sin(x+3*pi/4) = 0

The solutions to this equation are x = -3pi/4 + npi and x = pi/4 + n*pi, where n is any integer.

b) To solve sin(3x+3)=cos(x-1), we can rewrite it as sin(3x+3) - cos(x-1) = 0. Then, using the identity sin(a) - cos(b) = -sqrt(2)*sin(a + pi/4 - b), we have:

-sqrt(2)*sin(3x+3 + pi/4 - (x-1)) = 0

which simplifies to:

sin(2x+7*pi/4) = 0

The solutions to this equation are x = -7pi/8 + npi/2 and x = -5pi/8 + npi/2, where n is any integer.

a) To solve 8sin² x/2-3sin x-4=0, we can use the substitution y = sin(x/2):

8y^2 - 3(2y)/(1+y^2) - 4 = 0

Multiplying by (1+y^2), we get:

8y^4 - 6y^3 - 3y^2 + 6y - 4 = 0

This polynomial is difficult to solve exactly, but we can use numerical methods or approximations to find the solutions.

b) To solve 4sin² x/2-cos² x/2=1,5+sin x, we can use the identity cos²(x/2) = 1/2 + 1/2*cos(x), and substitute y = sin(x/2):

4y^2 - (1/2 + 1/2*cos(x)) - 1.5 - 2y = 0

Multiplying by 2, we get:

8y^2 - cos(x) - 5 - 4y = 0

We can use numerical methods or approximations to find the solutions.

3. To solve tg(x-15°)ctg(x+15°)=⅓, we can use the identity ctg(x) = 1/tg(x), and substitute y = tg(x):

y(y-2sqrt(3))/3 = 1

Multiplying by 3 and rearranging, we get:

y^2 - 2sqrt(3)*y - 3 = 0

Using the quadratic formula, we get:

y = sqrt(3) ± 2

Substituting back for tg(x), we have:

tg(x) = sqrt(3) + 2 or tg(x) = sqrt(3) - 2

The solutions to this equation are x = 15° + n180° + arctan(sqrt(3) + 2) and x = 15° + n180° + arctan(sqrt(3) - 2), where n is any integer.