Solve the given equation giving the general solution and all the values of theta?

Question

where 0 is greater than or equal to theta and less than or equal to 360 degrees.

4sec^2theta - 3 tan theta - 5 =0

and

b). find tan x if 2-3tan xcos^2x= cos^2x

Answer 1

Do you mean "where 0 is LESS THAN or equal to θ"?

Remember that tan^2(θ) + 1 = sec^2(θ).  So this is

4sec^2(θ) - 3tan(θ) - 5 = 0

4(tan^2(θ)+1) - 3tan(θ) - 5 = 0

4tan^2(θ) - 3tan(θ) - 1 = 0

Let y= tan(θ), and you get

4y^2 - 3y - 1 = 0

(4y + 1)(y - 1) = 0

y = -1/4, 1

So tan(θ) = -1/4, 1

This means θ = 45, 225, and the two values where tan(θ) = -1/4.

Answer 2

I am using angle A instead of theta

4sec^2A  - 3 tanA - 5 = 0

1+tan^2A = sec^2A

subsituting we get

4(1+tan^2A)-3tanA-5 = 0

4tan^2A-3tanA-5+4 = 0

4tan^2A-3tanA - 1 =0

put tan^A = y

4y^2-3y-1 =0

4y^2-4y+y-1 = 0

4y(y-1)+(y-1) = 0

ie (4y-1)(y-1) = 0

so y = -1/4   or y = 1

ie tan A = -1/4 or tan A =1

for tan A = .25 we get A as 166 or 346

for tan A = 1   we get A as 45  or 225

for the second part of the question do it similar way

Divide by cos^2X through out

we get

2sec^2X - 3tanX = 1

2tan^2X - 3tanX +2 = 1

2tan^2x-3tanX +1 =0

(2tanX-1)(tanX-1) = 0

tanX = 1/2 or 1

Answer 3

sec^2theta=1+tan^2 theta

==>

4tan^2theta - 3 tan theta - 1 =0

==>

tan theta=-1/4 or 1

==>

theta=arctan(-1/4)+Pi or arctan(-1/4)+2Pi

or Pi/4 or 5Pi/4.

tan xcos^2x=sinxcosx=sin2x,

cos^2x=(1+cos2x)/2

==>

3-6sin2x=cos2x.

notice that sin^2+cos^2=1,

it is easy to solve.

Answer 4

tan theta=1/4,-1