Solve the following equations for 0 < x < 2pi

A) sin^2x=2sinxcosx

B) 3tanx=cosx

Please help
Thankyou

A

x = 0 is a solution
also
sin x = 2 cos x
is
tan x = 2
that happens at x = 63.4 degrees and at x = 243.4 degrees
convert those to radians by multiplying by pi/180

B

3 (sin x/ cos x) = cos x
3 sin x = cos^2 x

3 sin x = (1 - sin^2 x)

sin^2 x + sin x -1 = 0

let z = sin x

z^2 + z - 1 = 0

z = [ -1 +/- sqrt (1+4) ]/2

z = [-1 +/- 2.24 ]/2

z = sin x = .618 or something too big to be a sin of anything

To solve the equations A) sin^2x = 2sinxcosx and B) 3tanx = cosx, we need to use trigonometric identities and algebraic manipulations. Here's how you can solve each equation step by step:

A) sin^2x = 2sinxcosx:

Step 1: Use the identity sin^2x = 1 - cos^2x to rewrite the equation:
1 - cos^2x = 2sinxcosx

Step 2: Rearrange the equation:
cos^2x + 2sinxcosx - 1 = 0

Step 3: Apply the identity 2sinxcosx = sin2x to obtain a quadratic equation in terms of sinx:
cos^2x + sin2x - 1 = 0

Step 4: Rewrite sin2x as 2sinxcosx using the double-angle identity:
cos^2x + 2sinxcosx - 1 = 0

Step 5: Factor the quadratic equation:
(cosx - 1)(cosx + 1) + 2sinxcosx - 1 = 0

Step 6: Simplify and rearrange terms:
cosx(cosx + 1) + (2sinx - 1)cosx = 0

Step 7: Factor out cosx from both terms:
cosx[(cosx + 1) + (2sinx - 1)] = 0

Step 8: Solve for cosx = 0:
cosx = 0, which implies x = pi/2 and x = 3pi/2 in the given domain 0 < x < 2pi.

Step 9: Solve for (cosx + 1) + (2sinx - 1) = 0:
cosx + 1 + 2sinx - 1 = 0
cosx + 2sinx = 0
Divide both sides by cosx:
1 + 2tanx = 0
2tanx = -1
tanx = -1/2
Using the unit circle or a trigonometric table, we can determine the solutions x where tanx = -1/2 within the given domain.

B) 3tanx = cosx:

Step 1: Divide both sides by cosx:
3tanx/cosx = 1

Step 2: Use the identity tanx = sinx/cosx:
3sinx/cos^2x = 1

Step 3: Rearrange the equation:
3sinx = cos^2x

Step 4: Use the identity tan^2x + 1 = sec^2x to rewrite the equation in terms of cosine:
3sinx = 1 - sin^2x

Step 5: Rearrange the equation:
sin^2x + 3sinx - 1 = 0

Step 6: Solve the quadratic equation for sinx using factoring, completing the square, or using the quadratic formula. The solutions for sinx will correspond to the solutions for x in the given domain 0 < x < 2pi.

Remember, it's essential to have a good understanding of trigonometric identities, algebraic manipulations, and solving equations to solve these types of problems.