How to solve these two trig equations

Question

a) tan² x + tanx =0

b) 2cos² x—3cosx +1=0

Answer 1

The key to solving these problems is to break them into two parts: 1)substitute tan x and cos x for y, and then solve the algebraic equation, 2) solve for x in tan x and cos x with the answers using the inverse tangent and cosine functions.

a) tan x^2 + tan x = 0

1)y = tan x

-> y^2 + y = 0
-> factor of y^2 + y = y(y + 1)
-> y(y + 1) = 0,
-> solving for y = {-1,0}

2)solve for x.

-> tan x = -1
-> tan x = 0

-> invtan(-1) = ?
-> invtan(0) = ?

x = invtan(0) = {0,pi}
x = invtan(-1) = {3*pi/4,7*pi/4}

X = {0,3*pi/4,pi,7*pi/4}



b)2 cos x^2 - 3 cos x + 1 = 0

1)y = cos x

->2y^2 - 3y + 1 = 0
->factor of 2y^2 - 3y + 1 = (2y - 1)(y - 1)
->(2y - 1)(y - 1) = 0
-> solving for y = {1/2, 1}

2)solve for x.

cos(x) = {1/2,1}

x = invcos(1/2)
x = invcos(1)

x = {pi/3,5*pi/3}
x = {0}

-> x = {0, pi/3, 5*pi/3}

All of the above answers can be added or subtracted by a multiple of 2*pi and will yield the same result.

for example, 0 +/- 2*n*pi where n = {0,1,...} gives {...-2*pi, 0, 2*pi, ...}

The cosine of all of these angles is still 1.

Answer 2

a) tanx(tanx+1) =0 ⇔ tanx =0 ⇒ x=k(180⁰) k being an integer; tanx = -1 ⇒ 135⁰ + k(180⁰)

 

b) 2cos²x-2cosx-1cosx+1=0 ⇔ 2cosx(cosx-1)-1(cosx-1)=0 ⇔ (cosx-1)(2cosx-1)=0 ⇒cosx=1 and x=k(180⁰)

    cosx=1/2 and x=60⁰ +k(360⁰) and x=300⁰+k(360⁰)