What is the general solution for sin(70)=1.15sin(x) ?

The general solution for sin(70)=1.15sin(x) is x=0.95640…+360n,x=180-0.95640…+360n

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sin(70)=1.15sin(x)

sin (7 0^{\circ}) = 1.15 sin (x)

x = 0.95640 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.95640 \dots + 36 0^{\circ} n

Radians

x = 0.95640 \dots + 2 πn, x = π - 0.95640 \dots + 2 πn

Solution steps

sin (7 0^{\circ}) = 1.15 sin (x)

Switch sides

1.15 sin (x) = sin (7 0^{\circ})

Divide both sides by

1.15

1.15 sin (x) = sin (7 0^{\circ})

Divide both sides by

1.15

\frac{1.15 sin ( x )}{1.15} = \frac{sin ( 7 0 ^{\circ} )}{1.15}

Simplify

sin (x) = \frac{sin ( 7 0 ^{\circ} )}{1.15}

sin (x) = \frac{sin ( 7 0 ^{\circ} )}{1.15}

Apply trig inverse properties

sin (x) = \frac{sin ( 7 0 ^{\circ} )}{1.15}

General solutions for

sin (x) = \frac{sin ( 7 0 ^{\circ} )}{1.15}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

x = arcsin (\frac{sin ( 7 0 ^{\circ} )}{1.15}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 7 0 ^{\circ} )}{1.15}) + 36 0^{\circ} n

x = arcsin (\frac{sin ( 7 0 ^{\circ} )}{1.15}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{sin ( 7 0 ^{\circ} )}{1.15}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.95640 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.95640 \dots + 36 0^{\circ} n

cos (x) = sin^{2} (x)

cos (b) = \frac{11}{14}

sin (x) - cos (x) = - 1, 0 \leq x \leq 2 π

cos (θ) = (\frac{1}{2})

sin^{2} (x) - cos^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π

What is the general solution for sin(70)=1.15sin(x) ?
The general solution for sin(70)=1.15sin(x) is x=0.95640…+360n,x=180-0.95640…+360n