Maths Please Help?

Question

Hello,

There are some questions that I am not sure how to tackle:

1) 3tanx = cosx (0 < x < pi)

2) tan^2x = tanx + 2 (0 < x < pi)

3) 1 + sinx = 2cos^2x (0 < x < pi)

The answers are apparently:

1) 0.308, 2.834

2) 1.107, [3(pi)] / 4, 2.356

3) (pi) / 6, [5(pi)] / 6

Solve the identity:

1) tanθ cosecθ = secθ

Thank you in advance.

Answer 1

1) 3tanx = cosx (0 < x < pi)

3sinx/cosx = cosx

3sinx =cos²x

3sinx =cos²x

Apply Pythagorean identity : sin²(u) + cos²(u) = 1.

3sinx =1-sin²x

sin²x+3sinx-1=0

Let sinx =t

t²+3t-1=0

Solve the  Quadratic equation:

sinx =0.303

sinx =sin (17.64)

General solution: x=nπ±(-1)^n17.64º

At n=0: x = (0)π+17.64º= 17.64º =0.3077 rad

At n=1: x = (1)π-17.64º= 3.14-0.3077 =2.832 rad

 

Answer 2

2) tan^2x = tanx + 2 (0 < x < pi)

tan²x = tanx + 2

tan²x - tanx - 2=0

Let tanx =t

t² - t - 2=0

Solve the  Quadratic equation:

t² -2 t +t- 2=0

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

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

t=2 or -1

tanx=2 or -1

tanx=-1

tanx =tan(3π/4)

General solution:x =nπ+3π/4

At n=0: x =3π/4

tanx=-1

tanx =tan(-π/4)

General solution:x =nπ-π/4

At n=1: x =π-π/4 =2.355

tanx=2

tanx =tan(63.43º)

General solution:x =nπ+63.43º

At n=0: x =63.43º =1.107 rad

Answer 3

1 + sinx = 2cos^2x (0 < x < pi)

Apply Pythagorean identity : sin²(u) + cos²(u) = 1.

1 + sinx =2(1 - sin²x)

1 + sinx =2 - 2sin²x

2sin²x+ sinx-1=0

Let sinx =t

2t²+t-1=0

Solve the  Quadratic equation:

2t²+2t-t-1=0

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

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

t =1/2 or -1

sinx =1/2

sinx =sin (π/6)

General solution: x=nπ±(-1)^n(π/6)

At n=0: x = (0)π+(-1)^0(π/6)= π/6

At n=1: x = (1)π+(-1)^1(π/6)=π-(π/6)= 5π/6 

sinx =-1 is not taken since angle is more than π

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Solve the identity:

1) tanθ cosecθ = secθ

Left hand identity :tanθ cosecθ

=(sinθ /cosθ)(1 /sinθ)

=(1 /cosθ)

=secθ

=Right hand identity