What is the general solution for sin(x)-cos(x)=1,0<= x<= 360 ?

The general solution for sin(x)-cos(x)=1,0<= x<= 360 is x=90,x=180

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sin(x)-cos(x)=1,0<= x<= 360

sin (x) - cos (x) = 1, 0 \leq x \leq 36 0^{\circ}

x = 9 0^{\circ}, x = 18 0^{\circ}

Solution steps

sin (x) - cos (x) = 1, 0 \leq x \leq 36 0^{\circ}

Rewrite using trig identities

2 sin (x - 4 5^{\circ}) = 1

Divide both sides by

2

sin (x - 4 5^{\circ}) = \frac{2}{2}

General solutions for

sin (x - 4 5^{\circ}) = \frac{2}{2}

x - 4 5^{\circ} = 4 5^{\circ} + 36 0^{\circ} n, x - 4 5^{\circ} = 13 5^{\circ} + 36 0^{\circ} n

Solve

x - 4 5^{\circ} = 4 5^{\circ} + 36 0^{\circ} n : x = 9 0^{\circ} + 36 0^{\circ} n

Solve

x - 4 5^{\circ} = 13 5^{\circ} + 36 0^{\circ} n : x = 36 0^{\circ} n + 18 0^{\circ}

x = 9 0^{\circ} + 36 0^{\circ} n, x = 36 0^{\circ} n + 18 0^{\circ}

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 9 0^{\circ}, x = 18 0^{\circ}

cos (x) = 0, 6

cos (x) = 0, 5

25 cos^{2} (x) = 9

cos (x) = \frac{2}{29}

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

What is the general solution for sin(x)-cos(x)=1,0<= x<= 360 ?
The general solution for sin(x)-cos(x)=1,0<= x<= 360 is x=90,x=180