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

The general solution for sin(x)+cos(x)=1,-360<= x<= 360 is x=0,x=90,x=360,x=-360,x=-270

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

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

x = 0, x = 9 0^{\circ}, x = 36 0^{\circ}, x = - 36 0^{\circ}, x = - 27 0^{\circ}

Solution steps

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

Solve

x + 4 5^{\circ} = 13 5^{\circ} + 36 0^{\circ} n : x = 36 0^{\circ} n + 9 0^{\circ}

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

Solutions for the range

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

x = 0, x = 9 0^{\circ}, x = 36 0^{\circ}, x = - 36 0^{\circ}, x = - 27 0^{\circ}

5 csc (θ) + 6 = 0

sin (x^{2}) = 1

cos (2 θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

sin (θ) = \frac{3}{10}, sin (θ + \frac{π}{6})

What is the general solution for sin(x)+cos(x)=1,-360<= x<= 360 ?
The general solution for sin(x)+cos(x)=1,-360<= x<= 360 is x=0,x=90,x=360,x=-360,x=-270