Question

sin3x*cos3x=-1/2
sin²2x-cos²2x=түбірдің астында3/2
cosx/4*sin пи/5-sinx/4*cos пи /5 =түбірдің астында -2/2

Answer 1

2. sin kb 2x- cos kb 2x= tubir 3/2

-(cos kb 2x-sin kb 2x)=tubir 3/2

cos 4x= -tubir 3/2

4x= -+(pi- arcos tubir 3/2)+ 2 pi*n. n tisti z ka

4x= -+(pi- pi/6)+ 2 pi*n. n tisti z ka

4x= -+5pi/6+ 2 pi*n. n tisti z ka

x= -+5pi/24+2 pi*n/4. n tisti z ka

x= -+5pi/24+ pi*n/2. n tisti z ka

Answer 2

1. sin 3x*cox3x=-1/2

1/2*2 sin 3x*cox3x=-1/2

bul gerde 1/2 men 2kiskaradi da ewtene ozgermeidi endi 1/2 tura beredi al 2sin 3x*cox3x=sin 6x-ti beredi

 1/2 sin 6x=- 1/2  koefisentten kutulu ywin tendiktin on jagina bolemiz

sin 6x=-1 kaladi osi jerden sin tin -1 degi manin jazganda

6x=(-1)^k+1arksin1+pi*k.k tisti z ka

6x=(-1)^k+1*pi/2+pi*k.k tisti z ka

x=(-1)^k+1*pi/12+pi*k/2 .k tisti z ka

 

Answer 3

cosx/4*sin пи/5-sinx/4*cos пи /5 =түбірдің астында -2/2

cosb*sina-sinb* cosa= sin (a-b)

sin(пи/5-x/4)=түбірдің астында -2/2

пи/5-x/4=(-1)^k+1arksin(2/2)+2pi*k.k tisti z ka

пи/5-x/4= (-1)k+1* pi/4+2pi*k.k tisti z ka

x/4=pi/5- (-1)k+1* pi/4+2pi*k.k tisti z ka

x=4 pi/5- (-1)k+1* pi+8pi*k.k tisti z ka