What is the general solution for cos(x+10)-cos(x+90)=1 ?

The general solution for cos(x+10)-cos(x+90)=1 is x=-50+360n+0.89125…,x=130+360n-0.89125…

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cos(x+10)-cos(x+90)=1

cos (x + 1 0^{\circ}) - cos (x + 9 0^{\circ}) = 1

x = - 5 0^{\circ} + 36 0^{\circ} n + 0.89125 \dots, x = 13 0^{\circ} + 36 0^{\circ} n - 0.89125 \dots

Solution steps

cos (x + 1 0^{\circ}) - cos (x + 9 0^{\circ}) = 1

Rewrite using trig identities

2 sin (4 0^{\circ}) sin (\frac{18 x + 90 0 ^{\circ}}{18}) = 1

Divide both sides by

2 sin (4 0^{\circ})

sin (\frac{18 x + 90 0 ^{\circ}}{18}) = \frac{1}{2 sin ( 4 0 ^{\circ} )}

Apply trig inverse properties

\frac{18 x + 90 0 ^{\circ}}{18} = arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )}) + 36 0^{\circ} n, \frac{18 x + 90 0 ^{\circ}}{18} = 18 0^{\circ} - arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )}) + 36 0^{\circ} n

Solve

\frac{18 x + 90 0 ^{\circ}}{18} = arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )}) + 36 0^{\circ} n : x = - 5 0^{\circ} + 36 0^{\circ} n + arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )})

Solve

\frac{18 x + 90 0 ^{\circ}}{18} = 18 0^{\circ} - arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )}) + 36 0^{\circ} n : x = 13 0^{\circ} + 36 0^{\circ} n - arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )})

x = - 5 0^{\circ} + 36 0^{\circ} n + arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )}), x = 13 0^{\circ} + 36 0^{\circ} n - arcsin (\frac{1}{2 sin ( 4 0 ^{\circ} )})

Show solutions in decimal form

x = - 5 0^{\circ} + 36 0^{\circ} n + 0.89125 \dots, x = 13 0^{\circ} + 36 0^{\circ} n - 0.89125 \dots

\frac{( cot ( x ) - 3 )}{( 2 sin ( x ) + 1 )} = 0

sin^{2} (x) + 3 sin (x) - 1 = 0

tan (x) - 3^{\frac{1}{2}} = 0

\frac{1 + ( 2 sin ( x ) )}{( cos ( x ) )} = 0

\frac{sin ^{2} ( a )}{2} = \frac{1 - cos ( a )}{2}

What is the general solution for cos(x+10)-cos(x+90)=1 ?
The general solution for cos(x+10)-cos(x+90)=1 is x=-50+360n+0.89125…,x=130+360n-0.89125…