What is the general solution for 4sin(x)+1=3csc(x) ?

The general solution for 4sin(x)+1=3csc(x) is x=(3pi)/2+2pin,x=0.84806…+2pin,x=pi-0.84806…+2pin

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4sin(x)+1=3csc(x)

4 sin (x) + 1 = 3 csc (x)

x = \frac{3 π}{2} + 2 πn, x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

Solution steps

4 sin (x) + 1 = 3 csc (x)

Subtract

3 csc (x)

from both sides

4 sin (x) + 1 - 3 csc (x) = 0

Rewrite using trig identities

1 + \frac{4}{csc ( x )} - 3 csc (x) = 0

Solve by substitution

csc (x) = - 1, csc (x) = \frac{4}{3}

csc (x) = - 1 : x = \frac{3 π}{2} + 2 πn

csc (x) = \frac{4}{3} : x = arccsc (\frac{4}{3}) + 2 πn, x = π - arccsc (\frac{4}{3}) + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arccsc (\frac{4}{3}) + 2 πn, x = π - arccsc (\frac{4}{3}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

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

csc (t) = \frac{1}{- \frac{2}{2}}

cos^{2} (x) = 1 - sin (x), 0 \leq x < 2 π

cos (2 x) - 3 sin (x) = 2, 0^{\circ} \leq x \leq 36 0^{\circ}

2 cos^{2} (x) + 3 cos (x) = 4 cos (x) + 1

What is the general solution for 4sin(x)+1=3csc(x) ?
The general solution for 4sin(x)+1=3csc(x) is x=(3pi)/2+2pin,x=0.84806…+2pin,x=pi-0.84806…+2pin