trigonometry questions solve

Question

1. tanx=2sinx from [0,2pi]
2. 2sin^x-cosx=1 from [0,2pi]

Answer 1

tanx = 2sinx

sinx/cosx = 2sinx

divide both sides by sinx

1/cosx = 2

cosx = 1/2

x = arccos.5 = 60 degrees or 300 degrees

also solve sinx = 0

x = arcsin0 = 0 or 360 degrees

x = 0, 60, 300 or 360

or

x = 0, pi/3, 5pi/3, 2pi

that's the solution to trig question 1

question 2 has a typo, misprint or mis-copied, as it doesn't make sense as written

maybe you meant 2sinx - cosx = 1. if so, then

you could solve it

two methods, graph it and find the x intercepts = the solutions

or

use trig & algebra

2sin^2(x) - cosx = 1

2(1-cos^2(x)) - cosx = 1

2 - 2cos^2(x) - cosx = 1

2cos^2(x) + cosx -1 = 0

let cosx = y

2y^2 +y -1 = 0

factor or use the quadratic formula

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

y = -1 or 1/2 = cosx

x = arccos-1 or arccos.5

x = pi or pi/3 or 5pi/3 radians

or

x = 60, 180 or 300 degrees

it is necessary to check the answers though, as you may introduce an extraneous solution, due to multiplying by a variable.

plug the "solutions" into the original equations and see if they work

2sin^2(60) -cos60 = 2(3/4) - 1/2 = 1. 60 degrees works

2sin^2(180) - cos180 = 0 +1 = 1. 180 works

2sin^2(300) - cos300 = 2(3/4) -1/2 = 1 300 works

no extraneous solutions

Answer 2

Solve the second part using the same technique. JonDT.com