Write the solution set on the interval [0,360) and to the nearst degree

Question

a) 4 sinx + 2 rad 3 = 0 

b) 6 cos ²x +7 cosx =5 

c)3 cos x= 2 

D) 4 sin²x=1

Answer 1

Hi Meg,

I am assuming rad refers to radical in a. For a while, I thought it was radians and couldn't solve the question. Please use sqrt or the radical sign instead.

 

a) 4sinx + 2sqrt(3) = 0

4sinx = -2sqrt(3)

sinx = -sqrt(3)/2

x = 240° and x = 300°

The solution set is x = {240, 300}

 

b) substitute u = cosx

6u^2 + 7u - 5 = 0

(2u - 1)(3u + 5) = 0

2u - 1 = 0  or 3u + 5 = 0

2u = 1           3u = -5

u = 1/2          u = -5/3

replace u by the original

cos x = 1/2

x = 60° and x = 300°

cos x = -5/3  this is invalid because cos x can not be less than -1 or greater than 1.

The solution set is x = {60, 300}

 

c) 3cosx = 2

cos x = 2/3

x = 48° and x = 312°

The solution set is x = {48, 312}

 

d) 4sin^2x = 1

sin^2x = 1/4

sinx = ±1/2

when sin x = 1/2, x = 30° and x = 150°

when sin x = -1/2, x = 210° and x = 330°

The solution set is x={30, 150, 210, 330}