I need help solving these equations for precalculus?

Question

1. (1+sqrt2 cosx)(1+tanx) = 0
2. tan^2x = sqrt3 tan x
can you also provide an explanation thanks

Answer 1

1. (1+sqrt2 cosx)(1+tanx) = 0

1+sqrt2 cosx=0  or 1+tanx=0

1+sqrt2 cosx=0 

sqrt2 cosx=-1

 cosx=-1/sqrt 2

x=3pi/4 ,5pi/4    0 ≦x ≦2pi

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1+tanx=0

tanx=-1

x=3pi/4,7pi/4

 

Total solutions x=3pi/4 ,5pi/4,7pi/4

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2)

tan^2x = sqrt3 tan x

tan^2x - sqrt3 tan x=0

tanx(tanx-sqrt3 )=0

tanx=0  or  tanx-sqrt3=0

tan x=0

x=0,pi,2pi

tanx-sqrt3=0

tanx=sqrt 3

x=pi/3,4pi/3

Total solutions x=0,pi,2pi,pi/3,4pi/3

Comments

tan x = 0 or tan x = √3.

• tan(x) = 0.

tan(x) = tan(0)

The genaral solution of tan(x) = tan(α) is x = nπ + α, where n is an integer.

x = nπ + 0.

The general solution is x = nπ .

• tan(x) = √3.

tan(x) = tan(π/3)

The genaral solution of tan(x) = tan(α) is x = nπ + α, where n is an integer.

x = nπ + π/3.

The general solution is x = nπ + π/3.

The general solutions of tan² x = √3tan x are x = nπ and x = nπ + π/3, where n is an integer.

Answer 2

(1).The trogonometric equation is [1 + √2 cos(x)][1 + tan(x)] = 0.

1 + √2 cos(x) = 0 and 1 + tan(x) = 0

cos(x) = - 1/√2 and tan(x) = - 1.

Equation 1: cos(x) = - 1/√2.

The function cos(x) has a period of 2π, first find all solutions in the interval [0, 2π).

The function cos(x) is negative in second and third quadrant.

In II quadrant (π/2 < θ < π),

cos(x) = - 1/√2 = cos(x) = - cos(π/4) = cos(π - π/4) = cos(3π/4) -------> x = 3π/4.

In III quadrant (π < θ < 3π/2),

cos(x) = - 1/√2 = cos(x) = - cos(π/4) = cos(π + π/4) = cos(5π/4) -------> x = 5π/4.

Finally, add multiples of 2π to each of these solutions to get the general form

x = 3π/4 + 2nπ and x = 5π/4 + 2nπ, where n is integer.

Equation 2 : tan(x) = - 1.

tan(x) = - tan(π/4)

tan(x) = tan(- π/4)

The genaral solution of tan(x) = tan(α) is x = nπ + α, where n is an integer.

x = nπ - π/4.

The general solutions of [1 + √2 cos(x)][1 + tan(x)] = 0 are x = 3π/4 + 2nπ, x = 5π/4 + 2nπ and x = nπ - π/4 where n is integer.