To solve the given equations, we can rewrite them in terms of sine and cosine functions:
1) sin^2(x) + 3sin(x) - 4 = 0 Let y = sin(x). The equation becomes: y^2 + 3y - 4 = 0 Factorizing, we get: (y + 4)(y - 1) = 0 So, either y + 4 = 0 or y - 1 = 0 Therefore, y = -4 or y = 1 Since y = sin(x), we have sin(x) = -4 or sin(x) = 1 Since sine values lie between -1 and 1, sin(x) = 1 is the only possible solution. Therefore, sin(x) = 1, which means x = π/2
2) cos^2(x) + 4cos(x) + 3 = 0 Let z = cos(x). The equation becomes: z^2 + 4z + 3 = 0 Factorizing, we get: (z + 3)(z + 1) = 0 So, either z + 3 = 0 or z + 1 = 0 Therefore, z = -3 or z = -1 Since z = cos(x), we have cos(x) = -3 or cos(x) = -1 Since cosine values lie between -1 and 1, there is no real solution to cos(x) = -3 or cos(x) = -1.
Therefore, the only solution to the system is x = π/2.
To solve the given equations, we can rewrite them in terms of sine and cosine functions:
1) sin^2(x) + 3sin(x) - 4 = 0
Let y = sin(x). The equation becomes:
y^2 + 3y - 4 = 0
Factorizing, we get:
(y + 4)(y - 1) = 0
So, either y + 4 = 0 or y - 1 = 0
Therefore, y = -4 or y = 1
Since y = sin(x), we have sin(x) = -4 or sin(x) = 1
Since sine values lie between -1 and 1, sin(x) = 1 is the only possible solution.
Therefore, sin(x) = 1, which means x = π/2
2) cos^2(x) + 4cos(x) + 3 = 0
Let z = cos(x). The equation becomes:
z^2 + 4z + 3 = 0
Factorizing, we get:
(z + 3)(z + 1) = 0
So, either z + 3 = 0 or z + 1 = 0
Therefore, z = -3 or z = -1
Since z = cos(x), we have cos(x) = -3 or cos(x) = -1
Since cosine values lie between -1 and 1, there is no real solution to cos(x) = -3 or cos(x) = -1.
Therefore, the only solution to the system is x = π/2.