Find the number of solutions for the given equation for 0° ≤ θ ≤ 360°.

sin2x - 2sinx - 3 = 0

Question

Find the number of solutions for the given equation for 0° ≤ θ ≤ 360°.

sin2x - 2sinx - 3 = 0

campbell_st · Answer

is the equation$$\sin^2 x - 2sinx - 3=0$$

anonymous · Answer

yes

anonymous · Answer

Its a quadratic in "sin(x)". Solve for the roots of the quadratic. Then, use inverse of sine to find x.

campbell_st · Answer

you need to factorise the quadratic and solve.... but there is only 1answer to the problem x = 270

anonymous · Answer

This is supposed to be a Trigonometric Equation.

anonymous · Answer

It is a qudratic equation with trigonometric function in its terms.

campbell_st · Answer

the factorisation is (sin x -3)(sinx + 1) = 0

anonymous · Answer

Since sin(x) can't be 3 ever, it only has one solution, which is sin(x) = -1 => x = 270 degrees.