Solve a 2sin x cot xcsc x0 xe027 b cos91 2 O360 cos Find al

Solve: a.) 2sin x + cot x-csc x=0 xe(0,27) b.) -cos9-1 2 [O.,360.) cos Find all solutions if 0 x

Solution

ans for (a) given 2sinx-cotx-cosecx =0

we know that cotx =cosx/sinx and cosecx =1/sinx

2sinx-cosx/sinx-1/sinx =0

taking sinx as common

(2sin^2x-cosx-1)/sinx =0

we can write sin^2x as 2cos^2x-1

(2(2cos^2x-1)-cosx-1) =0

-2cos^2x-cosx-1=0

let cosx=x

-2x^2-x-1 =0

(-2x+1)(x+1) =0

(-2cosx+1)(cosx+1) =0

cosx=1/2

now x=[/3,5/3]

if cosx =-1

x =

ans for (b)

we can solve this by using trigonometry identities

I am solving this problem by changing teta into \'t\'

given cos(t/2)-cos(t) =1

we know that cost =2cos^2(t/2) -1

now we get equation cos(t/2)-2cos^(t/2)+1 =1

cos(t/2) -2cos^(t/2) =0

take cos(t/2) as common

cos(t/2)(1-2cos^(t/2)) =0

Now by using trigonometric table of arcs and unit circle

there are two solutions

a.cos(t/2) =0

t/2 =/2 t =

t/2 =3/2 t =3 the value of 3pi is also equal to pi

b. cos(t/2) =1/2 t/2 =±/3

t/2 = /3 t = 2/3

t/2 =-/3 t =-2/3 or t/2 =5/3

t =10/3 =(4/3+2) =4/3

hence the values are ,2/3 ,4/3

Answer for part (b) a. given cos2x =1/2

we know that cos(/4) =1/2

cos2x =cos(/4)

2x = ±/4

x = ±/8

answer is x =/8 or x = -/8

Answer for part b (b) given sin2xcosx +cos2xsinx =1/2

we can solve this by using trigonometry formulas

(1) we know that sin(2x) =2sinxcosx and cos2x =2cos^2x-1

it is in the form of sinacosb +cosasinb =sin(a +b)

hera sina =sin2x and cosb =cosx

cosa =cos2x and sinb =sinx

sin2xcosx +cos2xsinx =sin(2x+x)

sin(3x) =1/2

sin(3x) =sin(30⁰)

3x=30⁰

we get x =10 degrees

and we also get

3x =150⁰

x=50⁰

answers for (0,360⁰) is 10⁰ ,50⁰