What is the general solution for sin(2θ)-cos(2θ)=1 ?

The general solution for sin(2θ)-cos(2θ)=1 is θ= pi/4+pin,θ=pin+pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2θ)-cos(2θ)=1

sin (2 θ) - cos (2 θ) = 1

θ = \frac{π}{4} + πn, θ = πn + \frac{π}{2}

Solution steps

sin (2 θ) - cos (2 θ) = 1

Rewrite using trig identities

2 sin (2 θ - \frac{π}{4}) = 1

Divide both sides by

2

sin (2 θ - \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (2 θ - \frac{π}{4}) = \frac{2}{2}

2 θ - \frac{π}{4} = \frac{π}{4} + 2 πn, 2 θ - \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

2 θ - \frac{π}{4} = \frac{π}{4} + 2 πn : θ = \frac{π}{4} + πn

Solve

2 θ - \frac{π}{4} = \frac{3 π}{4} + 2 πn : θ = πn + \frac{π}{2}

θ = \frac{π}{4} + πn, θ = πn + \frac{π}{2}

cos (θ) = \frac{1}{3}

tan (x) + \frac{1}{tan ( x )} = 2

4 csc (θ) - 5 = 3

csc (5 x) = - 1, 0 \leq x \leq 2 π

- 4 \cdot cos^{2} (x) + 3 = 0

What is the general solution for sin(2θ)-cos(2θ)=1 ?
The general solution for sin(2θ)-cos(2θ)=1 is θ= pi/4+pin,θ=pin+pi/2