What is the general solution for 90^2=60^2+70^2-2(60)(70)(cos(x)) ?

The general solution for 90^2=60^2+70^2-2(60)(70)(cos(x)) is x=1.52315…+2pin,x=2pi-1.52315…+2pin

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90^2=60^2+70^2-2(60)(70)(cos(x))

9 0^{2} = 6 0^{2} + 7 0^{2} - 2 (60) (70) (cos (x))

x = 1.52315 \dots + 2 πn, x = 2 π - 1.52315 \dots + 2 πn

Solution steps

9 0^{2} = 6 0^{2} + 7 0^{2} - 2 (60) (70) (cos (x))

Switch sides

6 0^{2} + 7 0^{2} - 2 \cdot 60 \cdot 70 cos (x) = 9 0^{2}

Multiply the numbers:

2 \cdot 60 \cdot 70 = 8400

6 0^{2} + 7 0^{2} - 8400 cos (x) = 9 0^{2}

6 0^{2} = 3600

3600 + 7 0^{2} - 8400 cos (x) = 9 0^{2}

7 0^{2} = 4900

3600 + 4900 - 8400 cos (x) = 9 0^{2}

Add the numbers:

3600 + 4900 = 8500

- 8400 cos (x) + 8500 = 9 0^{2}

9 0^{2} = 8100

- 8400 cos (x) + 8500 = 8100

Move

8500

to the right side

- 8400 cos (x) = - 400

Divide both sides by

- 8400

cos (x) = \frac{1}{21}

Apply trig inverse properties

x = arccos (\frac{1}{21}) + 2 πn, x = 2 π - arccos (\frac{1}{21}) + 2 πn

Show solutions in decimal form

x = 1.52315 \dots + 2 πn, x = 2 π - 1.52315 \dots + 2 πn

sin^{2} (x) + 3 sin (x) + 7 = 5

cos (2 x) = - 0.6, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

23 cos (\frac{θ}{2}) = - 3

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

3 cos (x) - sin (x) = 2

What is the general solution for 90^2=60^2+70^2-2(60)(70)(cos(x)) ?
The general solution for 90^2=60^2+70^2-2(60)(70)(cos(x)) is x=1.52315…+2pin,x=2pi-1.52315…+2pin