What is the general solution for 4cos(x-45)+7=10,0<= x<= 360 ?

The general solution for 4cos(x-45)+7=10,0<= x<= 360 is x=(180+4*0.72273…)/4 ,x=(180-4*0.72273…)/4

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4cos(x-45)+7=10,0<= x<= 360

4 cos (x - 4 5^{\circ}) + 7 = 10, 0^{\circ} \leq x \leq 36 0^{\circ}

x = \frac{18 0 ^{\circ} + 4 \cdot 0.72273 \dots}{4}, x = \frac{18 0 ^{\circ} - 4 \cdot 0.72273 \dots}{4}

Solution steps

4 cos (x - 4 5^{\circ}) + 7 = 10, 0^{\circ} \leq x \leq 36 0^{\circ}

Move

7

to the right side

4 cos (x - 4 5^{\circ}) = 3

Divide both sides by

4

cos (x - 4 5^{\circ}) = \frac{3}{4}

Apply trig inverse properties

x - 4 5^{\circ} = arccos (\frac{3}{4}) + 36 0^{\circ} n, x - 4 5^{\circ} = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

Solve

x - 4 5^{\circ} = arccos (\frac{3}{4}) + 36 0^{\circ} n : x = arccos (\frac{3}{4}) + 36 0^{\circ} n + 4 5^{\circ}

Solve

x - 4 5^{\circ} = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n : x = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n + 4 5^{\circ}

x = arccos (\frac{3}{4}) + 36 0^{\circ} n + 4 5^{\circ}, x = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n + 4 5^{\circ}

Solutions for the range

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

x = \frac{18 0 ^{\circ} + 4 arccos ( \frac{3}{4} )}{4}, x = \frac{18 0 ^{\circ} - 4 arccos ( \frac{3}{4} )}{4}

Show solutions in decimal form

x = \frac{18 0 ^{\circ} + 4 \cdot 0.72273 \dots}{4}, x = \frac{18 0 ^{\circ} - 4 \cdot 0.72273 \dots}{4}

6 cosh^{2} (x) + 4 sinh (x) = 7

53 tan (x) + 3 = 83 tan (x)

cos^{2} (x) = - 0.5

cos (x) = \frac{1.5}{4.272}

\frac{π}{12} = arcsin (\frac{x}{2})

What is the general solution for 4cos(x-45)+7=10,0<= x<= 360 ?
The general solution for 4cos(x-45)+7=10,0<= x<= 360 is x=(180+4*0.72273…)/4 ,x=(180-4*0.72273…)/4