What is the general solution for 4sin^2(x)+8cos(x)-7=0 ?

The general solution for 4sin^2(x)+8cos(x)-7=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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4sin^2(x)+8cos(x)-7=0

4 sin^{2} (x) + 8 cos (x) - 7 = 0

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

4 sin^{2} (x) + 8 cos (x) - 7 = 0

Rewrite using trig identities

- 3 - 4 cos^{2} (x) + 8 cos (x) = 0

Solve by substitution

cos (x) = \frac{1}{2}, cos (x) = \frac{3}{2}

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{3}{2} :

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

6 (cos (b))^{2} + cos (b) - 1 = 0

sin (θ) = \frac{1.33}{2.42}

2 tan (x) = 3 - tan^{2} (x), - 2 π \leq, x \leq π

- 25 = (29 cos (x)) \cdot 3.7 - 4.9 \cdot 3. 7^{2}

t, x = 5 (sin (t) + cos (t))

What is the general solution for 4sin^2(x)+8cos(x)-7=0 ?
The general solution for 4sin^2(x)+8cos(x)-7=0 is x= pi/3+2pin,x=(5pi)/3+2pin